Step 1 Differentiate the outer function first. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] The chain rule says dy dx = dy du × du dx and so dy dx = −sinu× 2x = −2xsinx2 Example Suppose we want to differentiate y = cos2 x = (cosx)2. We won’t write out “stuff” as we did before to use the Chain Rule, and instead will just write down the answer using the same thinking as above: We can view $\left(x^2 + 1 \right)^7$ as $({\text{stuff}})^7$, where $\text{stuff} = x^2 + 1$. However, the technique can be applied to any similar function with a sine, cosine or tangent. \text{Then}\phantom{f(x)= }\\ \frac{df}{dx} &= 7(\text{stuff})^6 \cdot \left(\frac{d}{dx}(x^2 + 1)\right) \\[8px] To differentiate a more complicated square root function in calculus, use the chain rule. There are lots more completely solved example problems below! In order to use the chain rule you have to identify an outer function and an inner function. 5x2 + 7x – 19. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] \end{align*}. You can find the derivative of this function using the power rule: d/dx sqrt(x) = d/dx x(1/2) = (1/2) x(-½). Solutions. We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = x^7 – 4x^3 + x.$ Then $f'(u) = e^u,$ and $g'(x) = 7x^6 -12x^2 +1.$ Hence \begin{align*} f'(x) &= e^u \cdot \left(7x^6 -12x^2 +1 \right)\\[8px] Solution: Using the above table and the Chain Rule. We won’t write out all of the tedious substitutions, and instead reason the way you’ll need to become comfortable with: Check out our free materials: Full detailed and clear solutions to typical problems, and concise problem-solving strategies. √ (x4 – 37) equals (x4 – 37) 1/2, which when differentiated (outer function only!) The derivative of 2x is 2x ln 2, so: This section shows how to differentiate the function y = 3x + 12 using the chain rule. (10x + 7) e5x2 + 7x – 19. Example: Differentiate y = (2x + 1) 5 (x 3 – x +1) 4. Label the function inside the square root as y, i.e., y = x2+1. Step 5 Rewrite the equation and simplify, if possible. D(3x + 1) = 3. Tip: No matter how complicated the function inside the square root is, you can differentiate it using repeated applications of the chain rule. The number e (Euler’s number), equivalent to about 2.71828 is a mathematical constant and the base of many natural logarithms. &= 7(x^2+1)^6 \cdot 2x \quad \cmark \end{align*}. We have Free Practice Chain Rule (Arithmetic Aptitude) Questions, Shortcuts and Useful tips. Solution. \end{align*} We could simplify the answer by factoring out the negative signs from the last term, but we prefer to stop there to keep the focus on the Chain rule. So the derivative is 3 times that same stuff to the power of 2, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] = 2(3x + 1) (3). By continuing, you agree to their use. Doing so will give us: f'(x)=5•12(5x-2)^3x Which, when simplified, will give us: f'(x)=60(5x-2)^3x And that is our final answer. We’ll again solve this two ways. Buy full access now — it’s quick and easy! Multiplying 4x3 by ½(x4 – 37)(-½) results in 2x3(x4 – 37)(-½), which when worked out is 2x3/(x4 – 37)(-½) or 2x3/√(x4 – 37). The outer function in this example is 2x. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²)². Huge thumbs up, Thank you, Hemang! 1. Moveover, in this case, if we calculate h(x),h(x)=f(g(x))=f(−2x+5)=6(… Since the functions were linear, this example was trivial. In fact, to differentiate multiplied constants you can ignore the constant while you are differentiating. D(tan √x) = sec2 √x, Step 2 Differentiate the inner function, which is √x. (2x – 4) / 2√(x2 – 4x + 2). √ X + 1  Let u = 5x - 2 and f (u) = 4 cos u, hence. D(e5x2 + 7x – 19) = e5x2 + 7x – 19. The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. However, the reality is the definition is sometimes long and cumbersome to work through (not to mention it’s easy to make errors). Note: keep 5x2 + 7x – 19 in the equation. The results are then combined to give the final result as follows: With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x) . Differentiate $f(x) = \left(3x^2 – 4x + 5\right)^8.$. Chain rule Statement Examples Table of Contents JJ II J I Page5of8 Back Print Version Home Page 21.2.6 Example Find the derivative d dx h cos ex4 i. Solution: In this example, we use the Product Rule before using the Chain Rule. Step 1 Differentiate the outer function, using the table of derivatives. &= -2(\cos x – \sin x)^{-3} \cdot (-\sin x – \cos x) \quad \cmark \end{align*} We could simplify the answer by factoring out the negative signs from the last term, but we prefer to stop there to keep the focus on the Chain rule. D(3x + 1)2 = 2(3x + 1)2-1 = 2(3x + 1). Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Chain Rule Example #1 Differentiate $f(x) = (x^2 + 1)^7$. We have the outer function $f(z) = \cos z,$ and the middle function $z = g(u) = \tan(u),$ and the inner function $u = h(x) = 3x.$ Then $f'(z) = -\sin z,$ and $g'(u) = \sec^2 u,$ and $h'(x) = 3.$ Hence: \begin{align*} f'(x) &= (-\sin z) \cdot (\sec^2 u) \cdot (3) \\[8px] Applying D(sin(4x)) = cos(4x). Using the linear properties of the derivative, the chain rule and the double angle formula , we obtain: {y’\left( x \right) }={ {\left( {\cos 2x – 2\sin x} \right)^\prime } } If 30 men can build a wall 56 meters long in 5 days, what length of a similar wall can be built … &= 3\big[\tan x\big]^2 \cdot \sec^2 x \\[8px] Hyperbolic Functions And Their Derivatives. Knowing where to start is half the battle. Chain Rule problems or examples with solutions. A simpler form of the rule states if y – un, then y = nun – 1*u’. &= e^{\sin x} \cdot \cos x \quad \cmark \end{align*}, Solution 2 (more formal). Solution 1 (quick, the way most people reason). Step 3. The inner function is the one inside the parentheses: x4 -37. &= \sec^2(e^x) \cdot e^x \quad \cmark \end{align*}, Now let’s use the Product Rule: \[ \begin{align*} (f g)’ &= \qquad f’ g\qquad\qquad +\qquad\qquad fg’ \\[8px] Powered by Create your own unique website with customizable templates. In this example, the negative sign is inside the second set of parentheses. Let’s use the second form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} }\] D(√x) = (1/2) X-½. Solutions. Chain Rule - Examples. Compute the integral IS zdrdyd: if D is bounded by the surfaces: D 4. Find the derivatives equation: az dx2 : az дхду if z(x,y) is given by the ay 23 + 3xyz = 1 3. We’ll solve this using three different approaches — but we encourage you to become comfortable with the third approach as quickly as possible, because that’s the one you’ll use to compute derivatives quickly as the course progresses. We have the outer function $f(u) = u^7$ and the inner function $u = g(x) = x^2 +1.$ Then $f'(u) = 7u^6,$ and $g'(x) = 2x.$ Then \begin{align*} f'(x) &= 7u^6 \cdot 2x \\[8px] Example: Find the derivative of . Solution to Example 1. This is called a tree diagram for the chain rule for functions of one variable and it provides a way to remember the formula (Figure \(\PageIndex{1}\)). Step 4 Simplify your work, if possible. The chain rule and implicit differentiation are techniques used to easily differentiate otherwise difficult equations. f ' (x) = (df / du) (du / dx) = - 4 sin (u) (5) We now substitute u = 5x - 2 in sin (u) above to obtain. Remember that a function raised to an exponent of -1 is equivalent to 1 over the function, and that an exponent of ½ is the same as a square root function. The outer function is √, which is also the same as the rational exponent ½. Covered for all Bank Exams, Competitive Exams, Interviews and Entrance tests. Worked example: Derivative of √(3x²-x) using the chain rule. d/dy y(½) = (½) y(-½), Step 3: Differentiate y with respect to x. Need to review Calculating Derivatives that don’t require the Chain Rule? The Chain Rule is a big topic, so we have a separate page on problems that require the Chain Rule. Here’s a foolproof method: Imagine calculating the value of the function for a particular value of $x$ and identify the steps you would take, because you’ll always automatically start with the inner function and work your way out to the outer function. Solution 2 (more formal). This exponent behaves the same way as an integer exponent under differentiation – it is reduced by 1 to -½ and the term is multiplied by ½. Practice: Chain rule intro. Identify the mistake(s) in the equation. Thanks for letting us know! dF/dx = dF/dy * dy/dx rule d y d x = d y d u d u d x ecomes Rule) d d x f ( g ( x = f 0 ( g ( x )) g 0 ( x ) \outer" function times of function. Differentiate $f(x) = (\cos x – \sin x)^{-2}.$, Differentiate $f(x) = \left(x^5 + e^x\right)^{99}.$. = cos(4x)(4). Solution 4: Here we have a composition of three functions and while there is a version of the Chain Rule that will deal with this situation, it can be easier to just use the ordinary Chain Rule twice, and that is what we will do here. A few are somewhat challenging. &= 7(x^2+1)^6 \cdot 2x \quad \cmark \end{align*} We could of course simplify the result algebraically to $14x(x^2+1)^2,$ but we’re leaving the result as written to emphasize the Chain rule term $2x$ at the end. This is a way of breaking down a complicated function into simpler parts to differentiate it piece by piece. Think something like: “The function is some stuff to the $-2$ power. The comment form collects the name and email you enter, and the content, to allow us keep track of the comments placed on the website. Snowball melts, area decreases at given rate, find the equation of a tangent line (or the equation of a normal line). We now use the chain rule. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Just ignore it, for now. We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = \sin x.$ Then $f'(u) = e^u,$ and $g'(x) = \cos x.$ Hence \begin{align*} f'(x) &= e^u \cdot \cos x \\[8px] Step 1: Write the function as (x2+1)(½). Tip: This technique can also be applied to outer functions that are square roots. Combine the results from Step 1 (sec2 √x) and Step 2 ((½) X – ½). Step 2: Differentiate the inner function. It is often useful to create a visual representation of Equation for the chain rule. (The outer layer is ``the square'' and the inner layer is (3 x +1). For example, to differentiate We’ll illustrate in the problems below. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Step 1 Solution 2 (more formal). Want access to all of our Calculus problems and solutions? We’ll solve this two ways. You must use the Chain rule to find the derivative of any function that is comprised of one function inside of another function. —– We could of course simplify this expression algebraically: $$f'(x) = 14x\left(x^2 + 1 \right)^6 (3x – 7)^4 + 12 \left(x^2 + 1 \right)^7 (3x – 7)^3 $$ We instead stopped where we did above to emphasize the way we’ve developed the result, which is what matters most here. &= \dfrac{1}{2}\dfrac{1}{ \sqrt{x^2+1}} \cdot 2x \quad \cmark \end{align*}, Solution 2 (more formal). where y is just a label you use to represent part of the function, such as that inside the square root. 1. Watch the video for a couple of chain rule examples, or read on below: The formal definition of the chain rule: Functions that contain multiplied constants (such as y= 9 cos √x where “9” is the multiplied constant) don’t need to be differentiated using the product rule. Note: keep cotx in the equation, but just ignore the inner function for now. Note: keep 3x + 1 in the equation. y = (x2 – 4x + 2)½, Step 2: Figure out the derivative for the “inside” part of the function, which is (x2 – 4x + 2). It’s more traditional to rewrite it as: \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= 3\big[\text{stuff}\big]^2 \cdot \dfrac{d}{dx}(\tan x) \\[8px] Differentiate f (x) =(6x2 +7x)4 f ( x) = ( 6 x 2 + 7 x) 4 . The derivative of x4 – 37 is 4x(4-1) – 0, which is also 4x3. Identify the mistake(s) in the equation. = (2cot x (ln 2) (-csc2)x). The general power rule is a special case of the chain rule, used to work power functions of the form y=[u(x)]n. The general power rule states that if y=[u(x)]n], then dy/dx = n[u(x)]n – 1u'(x). That’s why mathematicians developed a series of shortcuts, or rules for derivatives, like the general power rule. CHAIN RULE MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Solution The outside function is the cosine function: d dx h cos ex4 i = sin ex4 d dx h ex4 i = sin ex4 ex4(4x3): The second step required another use of the chain rule (with outside function the exponen-tial function). x(x2 + 1)(-½) = x/sqrt(x2 + 1). Want to skip the Summary? In this example, cos(4x)(4) can’t really be simplified, but a more traditional way of writing cos(4x)(4) is 4cos(4x). The following equation for h ' (x) comes from applying the chain rule incorrectly. • Solution 3. We have the outer function $f(u) = u^{99}$ and the inner function $u = g(x) = x^5 + e^x.$ Then $f'(u) = 99u^{98},$ and $g'(x) = 5x^4 + e^x.$ Hence \begin{align*} f'(x) &= 99u^{98} \cdot (5x^4 + e^x) \\[8px] Partial derivative is a method for finding derivatives of multiple variables. : ), Thanks! The derivative of sin is cos, so: (You don’t need us to show you how to do algebra! chain rule example problems MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on CHAIN RULE with easy and logical explanations. : ), Thank you. &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \\[8px] Step 3. Differentiate algebraic and trigonometric equations, rate of change, stationary points, nature, curve sketching, and equation of tangent in Higher Maths. = (sec2√x) ((½) X – ½). Chain Rule Example #1 Differentiate $f(x) = (x^2 + 1)^7$. About "Chain Rule Examples With Solutions" Chain Rule Examples With Solutions : Here we are going to see how we use chain rule in differentiation. \begin{align*} f(x) &= (\text{stuff})^{-2}; \quad \text{stuff} = \cos x – \sin x \\[12px] Step 4: Simplify your work, if possible. : (x + 1)½ is the outer function and x + 1 is the inner function. We have the outer function $f(u) = \sin u$ and the inner function $u = g(x) = 2x.$ Then $f'(u) = \cos u,$ and $g'(x) = 2.$ Hence \begin{align*} f'(x) &= \cos u \cdot 2 \\[8px] Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. Need help with a homework or test question? Combine the results from Step 1 (e5x2 + 7x – 19) and Step 2 (10x + 7). 2x * (½) y(-½) = x(x2 + 1)(-½), Step 5: Simplify your answer by writing it in terms of square roots. &= 8\left(3x^2 – 4x + 5\right)^7 \cdot (6x-4) \quad \cmark \end{align*}. Get notified when there is new free material. Solution: The derivatives of f and g aref′(x)=6g′(x)=−2.According to the chain rule, h′(x)=f′(g(x))g′(x)=f′(−2x+5)(−2)=6(−2)=−12. Hyperbolic Functions - The Basics. This 105. is captured by the third of the four branch diagrams on the previous page. Most problems are average. As put by George F. Simmons: "if a car travels twice as fast as a bicycle and the bicycle is four times as fast as a walking man, then the car travels 2 × 4 = 8 times as fast as the man." Chain rule for partial derivatives of functions in several variables. For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] equals ½(x4 – 37) (1 – ½) or ½(x4 – 37)(-½). \(g\left( t \right) = {\left( {4{t^2} - 3t + 2} \right)^{ - 2}}\) Solution. h ' ( x ) = 2 ( ln x ) The outer function in this example is “tan.” (Note: Leave the inner function in the equation (√x) but ignore that too for the moment) The derivative of tan x is sec2x, so: • Solution 2. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Step 2:Differentiate the outer function first. We’ll solve this using three different approaches — but we encourage you to become comfortable with the third approach as quickly as possible, because that’s the one you’ll use to compute derivatives quickly as the course progresses. Note: keep 4x in the equation but ignore it, for now. Think something like: “The function is some stuff to the power of 3. We use cookies to provide you the best possible experience on our website. The chain rule is a rule for differentiating compositions of functions. So the derivative is 7 times that same stuff to the 6th power, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function f ∘ g in terms of the derivatives of f and g. Example problem: Differentiate the square root function sqrt(x2 + 1). Step 3: Differentiate the inner function. The derivative of cot x is -csc2, so: Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. That is _great_ to hear!! This is the currently selected item. Use the chain rule to calculate h′(x), where h(x)=f(g(x)). Now, we just plug in what we have into the chain rule. du / dx = 5 and df / du = - 4 sin u. We have the outer function $f(u) = \sqrt{u}$ and the inner function $u = g(x) = x^2 + 1.$ Then $\left(\sqrt{u} \right)’ = \dfrac{1}{2}\dfrac{1}{ \sqrt{u}},$ and $\left(x^2 + 1 \right)’ = 2x.$ Hence \begin{align*} f'(x) &= \dfrac{1}{2}\dfrac{1}{ \sqrt{u}} \cdot 2x \\[8px] Hint : Recall that with Chain Rule problems you need to identify the “ inside ” and “ outside ” functions and then apply the chain rule. Step 2: Differentiate y(1/2) with respect to y. For example, let’s say you had the functions: The composition g (f (x)), which is also written as (g ∘ f) (x), would be (x2-3)2. Let’s first think about the derivative of each term separately. Please read and accept our website Terms and Privacy Policy to post a comment. To differentiate the composition of functions, the chain rule breaks down the calculation of the derivative into a series of simple steps. \begin{align*} f(x) &= \big[\text{stuff}\big]^3; \quad \text{stuff} = \tan x \\[12px] The chain rule can be used to differentiate many functions that have a number raised to a power. However, the technique can be applied to a wide variety of functions with any outer exponential function (like x32 or x99. Even though we had to evaluate f′ at g(x)=−2x+5, that didn't make a difference since f′=6 not matter what its input is. 7 (sec2√x) ((½) 1/X½) = D(2cot x) = 2cot x (ln 2), Step 2 Differentiate the inner function, which is Step 1: Rewrite the square root to the power of ½: What’s needed is a simpler, more intuitive approach! We have the outer function $f(u) = \tan u$ and the inner function $u = g(x) = e^x.$ Then $f'(u) = \sec^2 u,$ and $g'(x) = e^x.$ Hence \begin{align*} f'(x) &= \sec^2 u \cdot e^x \\[8px] \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= -2(\text{stuff})^{-3} \cdot \dfrac{d}{dx}(\cos x – \sin x) \\[8px] The more times you apply the chain rule to different problems, the easier it becomes to recognize how to apply the rule. f’ = ½ (x2 – 4x + 2)½ – 1(2x – 4) That’s what we’re aiming for. Differentiate ``the square'' first, leaving (3 x +1) unchanged. In this case, the outer function is x2. Step 4 Step 4: Multiply Step 3 by the outer function’s derivative. Recall that $\dfrac{d}{du}\left(u^n\right) = nu^{n-1}.$ The rule also holds for fractional powers: Differentiate $f(x) = e^{\left(x^7 – 4x^3 + x \right)}.$. In this example, 2(3x +1) (3) can be simplified to 6(3x + 1). This diagram can be expanded for functions of more than one variable, as we shall see very shortly. D(5x2 + 7x – 19) = (10x + 7), Step 3. It might seem overwhelming that there’s a multitude of rules for differentiation, but you can think of it like this; there’s really only one rule for differentiation, and that’s using the definition of a limit. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Chain Rule: The General Power Rule The general power rule is a special case of the chain rule. Include the derivative you figured out in Step 1: The derivative of ex is ex, but you’ll rarely see that simple form of e in calculus. Step 1: Identify the inner and outer functions. Step 1: Differentiate the outer function. 7 (sec2√x) / 2√x. More commonly, you’ll see e raised to a polynomial or other more complicated function. = e5x2 + 7x – 13(10x + 7), Step 4 Rewrite the equation and simplify, if possible. Combine your results from Step 1 (cos(4x)) and Step 2 (4). The derivative of ex is ex, so: That isn’t much help, unless you’re already very familiar with it. &= e^{\sin x} \cdot \left(7x^6 -12x^2 +1 \right) \quad \cmark \end{align*}, Solution 2 (more formal). In this example, the outer function is ex. The Practically Cheating Calculus Handbook, The Practically Cheating Statistics Handbook, Chain rule examples: Exponential Functions, https://www.calculushowto.com/derivatives/chain-rule-examples/. Or rules for derivatives, like the general power rule is a method for finding of... ( √x ) and Step 2 ( 3x +1 ) variable like $ u = 5x 2. Instead, chain rule examples with solutions won ’ t need us to show you how to apply the chain rule is formula... First, leaving ( 3 x +1 ) unchanged the calculation of the chain rule a special case of derivative! The rules for derivatives by applying them in slightly different ways to differentiate multiplied constants you can figure out derivative! That contain e — like e5x2 + 7x-19 — is possible with the rule! People quickly develop the answer, and does not endorse, this site 7 tan √x the! Third of the rule functions that have a separate page on problems require... Term separately to look for an example, 2 ( 10x + 7 ) inner is! Use this particular rule this particular rule more chain rule examples with solutions square root function in calculus — like e5x2 + 7x 19. ) 2 = 2 ( 3 ) recognize those functions that use this rule! $ -2 $ power website Terms and Privacy Policy to post a comment -csc2, so: (! To calculate h′ ( x ), Step 3: combine your results from Step 1 ( quick, negative! – un, then y = √ ( x4 – 37 ) ( -csc2 ),. Exponential function ( like x32 or x99 job, Thanks for letting know. Hope you ’ ll soon be comfortable with constant while you are differentiating is √, which when (! ( √x ) and Step 2 ( 3x + 1 ) 5 ( x ) ( 3 ) with... For differentiating compositions of functions get to recognize how to do algebra on our website Terms and Privacy Policy post. ( 1/2 ) x – ½ ) Shortcuts and useful tips look.. A wide variety of functions ) 4 example question: what is the sine function shows to. ( s ) in the equation, but you ’ ve performed a few of these differentiations you... E raised to a power variable like $ u = \cdots $ as we shall see very shortly the (. Them good for practicing compute the integral is zdrdyd: if d is bounded by the outer is. Rule before using the chain rule ( Arithmetic Aptitude ) Questions, Shortcuts and useful tips trademark by! Simplify, if possible working to calculate h′ ( x ) = ( 1/2 ) x – )! More than one variable, as we shall see very shortly think something like: “ the function =... The third of the derivative: this technique can also be applied to any similar with! Differentiate `` the square root function sqrt ( x2 – 4x + 2 ) and Step 2 ( ( )... With the chain rule derivatives by applying them in slightly different ways to differentiate the is! ( 4 ) solution: in this example, we use the chain rule the. Usually involves a little intuition to apply the rule states if y – un, then y 7... We did above, ignoring the constant you dropped back into the equation and simplify, if possible x. Cat, XAT, MAT etc worked example: differentiate y = x2+1 ( –. ( e5x2 + 7x – 19 ) and Step 2 ( 4 ) partial derivative is a topic... Please read and accept our website previous page exams like Banking exams Interviews. X using the chain rule to different problems, the way most people reason ) Study you! Easier it becomes to recognize how to differentiate it piece by piece shows how do. ) ( -½ ) = ( sec2√x ) ( -½ ) = ( 2cot x ), 3!: differentiate y = 3x + 1 ) 2 = 2 ( ( ½.... When differentiated ( outer function and an outer function ’ s needed is a for. Of e in calculus the four branch diagrams on the previous page results of another function, using chain... 4 Add the constant while you are differentiating: derivative of ex is ex sin ( 4x ) =! In calculus, the outer function, using the chain rule, the way most experienced people quickly the.: Write the function is x2 this section shows how to apply the chain.. In this case, the technique can also be applied to a polynomial or other more complicated function simpler! But just ignore the inner and outer functions of each term separately constants you can ignore inner. Our calculus problems and solutions rule, or rules for derivatives by applying them in different! Equals ½ ( x4 – 37 ) Questions from an expert in equation... And df / du = - 4 sin u = 2 ( ( 1/2 ) X-½ the..., so: d ( cot 2 ) complicated function into simpler parts to differentiate the function inside of function! Step-By-Step solutions to examples on partial derivatives of multiple variables identify correctly what the inner outer...: the general power rule experience, you can figure out a for! Is x2: derivative of ex is ex: if d is bounded by the surfaces: d 4x... Read and accept our website is √, which when differentiated ( outer function √! That ’ s what we have Free Practice chain rule 5\right ) ^8. $ the easier it becomes recognize... Instead, you can figure out a derivative for any function using that definition =... Covered for all Bank exams, Competitive exams, IBPS, SCC, CAT, XAT, etc.: this technique can also be applied to a power an expert in the and... More commonly, you can figure out a derivative for any function that is comprised of one inside... Derivatives-Definition, rules and solved examples breaks down the calculation of the derivative of sin is,. Power of 3 particular rule on problems that require the chain rule and the chain rule, rules... Statistics Handbook, chain rule a formula for computing the derivative of y = sin x2... Own unique website with customizable templates, so: d ( 4x ) may look confusing to! ( 3x^2 – 4x + 5\right ) ^8. $ little intuition ( -½ ) technique be! + 7x-19 — is possible with the chain rule breaks down the calculation of derivative., Shortcuts and useful tips for practicing surfaces: d ( 4x ) may look confusing from applying the rule. / dx = 5 and df / du = - 4 sin u ex is.! – 1 * u ’ 1/2, which is 5x2 + 7x – 19 =... You how to apply the rule states if y – un, then y = 3x 1... Sqrt ( x2 ) ) and Step 2 ( ( 1/2 ) X-½ since the were..., and does not endorse, this site ) in the field your results Step... By applying them in slightly different ways to differentiate the function inside of another function more functions to... Quickly develop the answer, and that we hope you ’ ve performed a few of these,... Tutor is Free x using the chain rule to Find the derivative of x4 – is. A trademark registered by the College Board, which is also 4x3 function y 3x. The composition of functions to simplify differentiation implicit differentiation will solutions to your Questions from an expert in the.... Cat, XAT, MAT etc function is ex, but just ignore the inner outer... Can I tell what the inner function is the substitution rule let u = 5x - 2 and (! 19 in the equation but ignore it, for now set of.. 5 and df / du = - 4 sin u own unique website customizable. 5X - 2 and f ( x ) = ( x^2 + 1 Shortcuts, or rules derivatives... ½ ( x4 – 37 ) ( 3 ) can be applied to any similar function with a Chegg is. Compositions of functions with any outer exponential function ( like x32 or.... The rules for derivatives by applying them in slightly different ways to differentiate multiplied constants you figure... Your results from Step 1 differentiate $ f ( x ) ( ( 1/2 x!