Using the chain rule and the quotient rule, \begin{equation} \frac{dy}{dx}=\frac{\sqrt{x^4+4}(1)-x\frac{d}{dx}\left(\sqrt{x^4+4}\right)}{\left(\sqrt{x^4+4}\right)^2}=\frac{\sqrt{x^4+4}(1)-x\left(\frac{2 x^3}{\sqrt{4+x^4}}\right)}{\left(\sqrt{x^4+4}\right)^2} \end{equation} which simplifies to \begin{equation} \frac{dy}{dx}=\frac{4-x^4}{\left(4+x^4\right)^{3/2}} \end{equation} as desired. Example. Let $f$ be a function for which $f(2)=-3$ and $$ f'(x)=\sqrt{x^2+5}. Evaluating Limits Analytically (Using Limit Theorems) [Video], Intuitive Introduction to Limits (The Behavior of a Function) [Video], Related Rates (Applying Implicit Differentiation), Numerical Integration (Trapezoidal and Simpson’s), Integral Definition (The Definite Integral), Indefinite Integrals (What is an antiderivative? Solution. Show that $$\frac{d}{d\theta }(\sin \theta {}^{\circ})=\frac{\pi }{180}\cos \theta .$$ What do you think is the importance of the exercise? On the other hand, simple basic functions such as the fifth root of twice an input does not fall under these techniques. To simplify the set-up, let’s assume that \(\mathbf g:\R\to \R^n\) and \(f:\R^n\to \R\) are both functions of class \(C^1\). ), Calculus (Start Here) – Enter the World of Calculus, Continuous (It’s Meaning and Applications), Derivative Definition (The Derivative as a Function), Derivative Examples (The Role of the Derivative), Find the Limit (Techniques for Finding Limits), First Derivative Test (and Curve Sketching), Horizontal Asymptotes and Vertical Asymptotes, Implicit Differentiation (and Logarithmic Differentiation), L ‘Hopital’s Rule and Indeterminate Forms, Limit Definition (Precise Definition of Limit), Choose your video style (lightboard, screencast, or markerboard). Define \(\phi = f\circ \mathbf g\). Using the chain rule, \begin{equation} \frac{d}{d x}f'[f(x)] =f” [ f(x)] f'(x) \end{equation} which is the second derivative evaluated at the function multiplied by the first derivative; while, \begin{equation} \frac{d}{d x}f [f'(x)]=f'[f'(x)]f”(x) \end{equation} is the first derivative evaluated at the first derivative multiplied by the second derivative. The proof of this theorem uses the definition of differentiability of a function of two variables. Assume for the moment that () does not equal () for any x near a. Sort by: Top Voted. In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. By the chain rule $$ g'(x)=f'(3x-1)\frac{d}{dx}(3x-1)=3f'(3x-1)=\frac{3}{(3x-1)^2+1}. 0�9���|��1dV Exercise. Under certain conditions, such as differentiability, the result is fantastic, but you should practice using it. f (z) = √z g(z) = 5z −8 f ( z) = z g ( z) = 5 z − 8. then we can write the function as a composition. Given: Functions and . If fis di erentiable at P, then there is a constant M 0 and >0 such that if k! Chain rule is a formula which is the same in standard and non-standard analysis. Next lesson. Let $u$ be a differentiable function of $x.$ Use $|u|=\sqrt{u^2}$ to prove that $$\frac{d}{dx}(|u| )=\frac{u’ u}{|u|} $$ when $u\neq 0.$ Use the formula to find $h’$ given $h(x)=x|2x-1|.$. Included Quiz Questions. Solution. By using the chain rule we determine, \begin{align} f'(x) & = \frac{\sqrt{2x-1}(1)-x\frac{d}{dx}\left(\sqrt{2x-1}\right)}{\left(\sqrt{2x-1}\right)^2} \\ & =\frac{\sqrt{2x-1}(1)-x \left(\frac{1}{\sqrt{-1+2 x}}\right)}{\left(\sqrt{2x-1}\right)^2} \end{align} which simplifies to $$ f'(x)=\frac{-1+x}{(-1+2 x)^{3/2}}. PQk< , then kf(Q) f(P)k> Suppose that f is differentiable at the point \(\displaystyle P(x_0,y_0),\) where \(\displaystyle x_0=g(t_0)\) and … Solution. $$ Also, by the chain rule \begin{align} h'(x) & = f’\left(\frac{1}{x}\right)\frac{d}{dx}\left(\frac{1}{x}\right) \\ & =-f’\left(\frac{1}{x}\right)\left(\frac{1}{x^2}\right) \\ & =\frac{-1}{\left(\frac{1}{x} \right)^2 + 1} \left(\frac{1}{x^2}\right) \\ & =\frac{-1}{x^2+1}. To prove: wherever the right side makes sense. If $y$ is a differentiable function of $u,$ $u$ is a differentiable function of $v,$ and $v$ is a differentiable function of $x,$ then $$ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}. Using the chain rule, \begin{align} \frac{dy}{dx}&=\cos \sqrt[3]{x}\frac{d}{dx}\left(\sqrt[3]{x}\right)+\frac{1}{3}(\sin x)^{-2/3}\frac{d}{dx}(\sin x) \\ & =\frac{1}{3 x^{2/3}}\cos \sqrt[3]{x}+\frac{\cos x}{3(\sin x)^{2/3}}. And, if you've been following some of the videos on "differentiability implies continuity", and what happens to a continuous function as our change in x, if x is our independent variable, as that approaches zero, how the change in our function approaches zero, then this proof is actually surprisingly straightforward, so let's just get to it, and this is just one of many proofs of the chain rule. Many undergraduate calculus texts assert the following proof of the chain rule. let t = 1 + x² therefore, y = t³ dy/dt = 3t² dt/dx = 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 Derivative rules review. Solution. In Calculus, a Quotient rule is similar to the product rule. 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). $$. Note that now in terms of this new variable, we need to prove: ddxf(u)=f′(u)dudx{\frac{d}{dx}f(u) = f'(u)\frac{du}{dx}} dxd​f(u)=f′(u)dxdu​ Since u = g(x), it is obviously a function of x; and since we have assumed g(x) to be differentiable, u will also be differentiable. PQk: Proof. Solution. Find the derivative of the function \begin{equation} g(x)=\left(\frac{3x^2-2}{2x+3}\right)^3. \end{equation} Therefore, \begin{equation} g'(2)=2(2) f\left(\frac{2}{2-1}\right)+2^2f’\left(\frac{2}{2-1}\right)\left(\frac{-1}{(2-1)^2}\right)=-24. As in single variable calculus, there is a multivariable chain rule. This proof uses the following fact: Assume , and . Specifically, it allows us to use differentiation rules on more complicated functions by differentiating the inner function and outer function separately. Example. This section gives plenty of examples of the use of the chain rule as well as an easily understandable proof of the chain rule. The gradient is one of the key concepts in multivariable calculus. Find the derivative of the following functions.$(1) \quad \displaystyle r=-(\sec \theta +\tan \theta )^{-1}$$(2) \quad \displaystyle y=\frac{1}{x}\sin ^{-5}x-\frac{x}{3}\cos ^3x$$(3) \quad \displaystyle y=(4x+3)^4(x+1)^{-3}$$(4) \quad \displaystyle y=(1+2x)e^{-2x}$$(5) \quad \displaystyle h(x)=x \tan \left(2 \sqrt{x}\right)+7$$(6) \quad \displaystyle g(t)=\left(\frac{1+\cos t}{\sin t}\right)^{-1}$$(7) \quad \displaystyle q=\sin \left(\frac{t}{\sqrt{t+1}}\right)$$(8) \quad \displaystyle y=\theta ^3e^{-2\theta }\cos 5\theta $$(9) \quad \displaystyle y=(1+\cos 2t)^{-4}$$(10) \quad \displaystyle y=\left(e^{\sin (t/2)}\right)^3$$(11) \quad \displaystyle y=\left(1+\tan ^4\left(\frac{t}{12}\right)\right)^3$$(12) \quad \displaystyle y=4 \sin \left(\sqrt{1+\sqrt{t}}\right)$$(13) \quad \displaystyle y=\frac{1}{9}\cot (3x-1)$$(14) \quad \displaystyle y=\sin \left(x^2e^x\right)$$(15) \quad \displaystyle y=e^x \sin \left(x^2e^x\right)$, Exercise. 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