Let u = cosx so that y = u2 It follows that du dx = −sinx dy du = 2u Then dy dx = dy du × du dx = 2u× −sinx = −2cosxsinx Example Suppose we wish to diﬀerentiate y = (2x− 5)10. 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. = cos(4x)(4). Step 2: Differentiate the inner function. : ). 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 $\text{stuff} = x^2 + 1$. Let u = 5x - 2 and f (u) = 4 cos u, hence. Let f(x)=6x+3 and g(x)=−2x+5. Combine the results from Step 1 (sec2 √x) and Step 2 ((½) X – ½). D(√x) = (1/2) X-½. This video gives the definitions of the hyperbolic functions, a rough graph of three of the hyperbolic functions: y = sinh x, y = cosh x, y = tanh x &= \cos(2x) \cdot 2 \quad \cmark \end{align*}, Solution 2. 2x * (½) y(-½) = x(x2 + 1)(-½), Step 5: Simplify your answer by writing it in terms of square roots. 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. 1. &= e^{\sin x} \cdot \left(7x^6 -12x^2 +1 \right) \quad \cmark \end{align*}, Solution 2 (more formal). Differentiate $f(x) = \left(3x^2 – 4x + 5\right)^8.$. This diagram can be expanded for functions of more than one variable, as we shall see very shortly. Chain Rule problems or examples with solutions. 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*}} = (2cot x (ln 2) (-csc2)x). It’s more traditional to rewrite it as: 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. Then. For instance, $\left(x^2+1\right)^7$ is comprised of the inner function $x^2 + 1$ inside the outer function $(\boxed{\phantom{\cdots}})^7.$ As another example, $e^{\sin x}$ is comprised of the inner function $\sin x$ inside the outer function $e^{\boxed{\phantom{\cdots}}}.$ As yet another example, $\ln{(t^3 – 2t^2 +5)}$ is comprised of the inner function $t^3 – 2t^2 +5$ inside the outer function $\ln(\boxed{\phantom{\cdots}}).$ Since each of these functions is comprised of one function inside of another function — known as a composite function — we must use the Chain rule to find its derivative, as shown in the problems below. Solution: Using the above table and the Chain Rule. Show Solution. Click HERE for a real-world example of 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*}} AP® is a trademark registered by the College Board, which is not affiliated with, and does not endorse, this site. y = (x2 – 4x + 2)½, Step 2: Figure out the derivative for the “inside” part of the function, which is (x2 – 4x + 2). 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. Moveover, in this case, if we calculate h(x),h(x)=f(g(x))=f(−2x+5)=6(… We’re glad to have helped! • Solution 1. Differentiate the square'' first, leaving (3 x +1) unchanged. 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). Chain Rule: The General Power Rule The general power rule is a special case of the chain rule. : (x + 1)½ is the outer function and x + 1 is the inner function. &= 99\left(x^5 + e^x\right)^{98} \cdot \left(5x^4 + e^x\right) \quad \cmark \end{align*}, Solution 2. Most problems are average. √ X + 1  When you apply one function to the results of another function, you create a composition of functions. A few are somewhat challenging. The chain rule in calculus is one way to simplify differentiation. 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. Step 5 Rewrite the equation and simplify, if possible. The following equation for h ' (x) comes from applying the chain rule incorrectly. The following chain rule examples show you how to differentiate (find the derivative of) many functions that have an “ inner function ” and an “ outer function.” For an example, take the function y = √ (x 2 – 3). No other site explains this nice. You can find the derivative of this function using the power rule: equals ½(x4 – 37) (1 – ½) or ½(x4 – 37)(-½). Hint : Recall that with Chain Rule problems you need to identify the “ inside ” and “ outside ” functions and then apply the chain rule. D(3x + 1)2 = 2(3x + 1)2-1 = 2(3x + 1). Solution 2 (more formal) . Technically, you can figure out a derivative for any function using that definition. We have $y = u^7$ and $u = x^2 +1.$ Then $\dfrac{dy}{du} = 7u^6,$ and $\dfrac{du}{dx} = 2x.$ Hence \begin{align*} \dfrac{dy}{dx} &= 7u^6 \cdot 2x \8px] Your first 30 minutes with a Chegg tutor is free! We have the outer function f(u) = u^8 and the inner function u = g(x) = 3x^2 – 4x + 5. Then f'(u) = 8u^7, and g'(x) = 6x -4. Hence \begin{align*} f'(x) &= 8u^7 \cdot (6x – 4) \\[8px] &= -2(\cos x – \sin x)^{-3} \cdot (-\sin x – \cos x)\quad \cmark \\[8px] 7 (sec2√x) ((½) X – ½) = Example: Differentiate y = (2x + 1) 5 (x 3 – x +1) 4. So the derivative is -2 times that same stuff to the -3 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)}} 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. Get complete access: LOTS of problems with complete, clear solutions; tips & tools; bookmark problems for later review; + MORE! Note that I’m using D here to indicate taking the derivative. Example question: What is the derivative of y = √(x2 – 4x + 2)? \begin{align*} f(x) &= \big[\text{stuff}\big]^3; \quad \text{stuff} = \tan x \12px] In this example, cos(4x)(4) can’t really be simplified, but a more traditional way of writing cos(4x)(4) is 4cos(4x). Step 2: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*}} That’s what we’re aiming for. 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] \end{align*}. 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] The more times you apply the chain rule to different problems, the easier it becomes to recognize how to apply the rule. Tip You can also use this rule to differentiate natural and common base 10 logarithms (D(ln x) = (1/x) and D(log x) = (1/x) log e. Multiplied constants add another layer of complexity to differentiating with the chain rule. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. h ' ( x ) = 2 ( ln x ) 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. \begin{align*} f(x) &= (\text{stuff})^{-2}; \quad \text{stuff} = \cos x – \sin x \\[12px] d/dx sqrt(x) = d/dx x(1/2) = (1/2) x(-½). Step 4 Rewrite the equation and simplify, if possible. 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. Step 3: Combine your results from Step 1 2(3x+1) and Step 2 (3). The Practically Cheating Calculus Handbook, The Practically Cheating Statistics Handbook, Chain rule examples: Exponential Functions, https://www.calculushowto.com/derivatives/chain-rule-examples/. In this example, no simplification is necessary, but it’s more traditional to write the equation like this: A simpler form of the rule states if y – un, then y = nun – 1*u’. $$g\left( t \right) = {\left( {4{t^2} - 3t + 2} \right)^{ - 2}}$$ Solution. Applying CHAIN RULE MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Watch the video for a couple of chain rule examples, or read on below: The formal definition of the chain rule: \text{Then}\phantom{f(x)= }\\ \frac{df}{dx} &= 7(\text{stuff})^6 \cdot \left(\frac{d}{dx}(x^2 + 1)\right) \\[8px] &= \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] Example: Find d d x sin( x 2). &= -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. \[ \bbox[10px,border:2px solid blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)} } Even though few people admit it, almost everyone thinks along the lines of the informal approach in the blue boxes above. Add the constant you dropped back into the equation. The problems below combine the Product rule and the Chain rule, or require using the Chain rule multiple times. Now, we just plug in what we have into the chain rule. Huge thumbs up, Thank you, Hemang! Include the derivative you figured out in Step 1: In this example, 2(3x +1) (3) can be simplified to 6(3x + 1). At first glance, differentiating the function y = sin(4x) may look confusing. D(2cot x) = 2cot x (ln 2), Step 2 Differentiate the inner function, which is In this case, the outer function is x2. (a) f(x;y) = 3x+ 4y; @f @x = 3; @f @y = 4. \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. Step 4: Simplify your work, if possible. We have Free Practice Chain Rule (Arithmetic Aptitude) Questions, Shortcuts and Useful tips. What’s needed is a simpler, more intuitive approach! With some experience, you won’t introduce a new variable like $u = \cdots$ as we did above. d/dy y(½) = (½) y(-½), Step 3: Differentiate y with respect to x. Since the functions were linear, this example was trivial. 7 (sec2√x) / 2√x. 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. The second is more formal. x(x2 + 1)(-½) = x/sqrt(x2 + 1). The number e (Euler’s number), equivalent to about 2.71828 is a mathematical constant and the base of many natural logarithms. Therefore sqrt(x) differentiates as follows: Step 3. We now use the chain rule. Combine your results from Step 1 (cos(4x)) and Step 2 (4). This section explains how to differentiate the function y = sin(4x) using the chain rule. dF/dx = dF/dy * dy/dx du / dx = 5 and df / du = - 4 sin u. Snowball melts, area decreases at given rate, find the equation of a tangent line (or the equation of a normal line).

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