Quiz 4 - Chain rule and implicit differentiation

Problem 1

Question

Find the derivative $\frac{dy}{dx}$ for:

a) $y=\sqrt{x^2+{\sec }^2(x)}$

b) $y={\log }_5(\csc (x))$

c) $y={\tan }^2(\sin (3x))$

Solution

a) $$\begin{array}{rl}y& =\sqrt{x^2+{\sec }^2(x)}={(x^2+{\sec }^2(x))}^{1/2}\\ \frac{dy}{dx}& =\frac{1}{2}{(x^2+{\sec }^2(x))}^{-1/2}\times \frac{d}{dx}(x^2+{\sec }^2(x))\\ & =\frac{1}{2}{(x^2+{\sec }^2(x))}^{-1/2}\times (2x+2\sec (x)\times \frac{d}{dx}\sec (x))\\ & =\frac{1}{2}{(x^2+{\sec }^2(x))}^{-1/2}\times (2x+2{\sec }^2(x)\tan (x))\\ \boxed{{\displaystyle \frac{dy}{dx}=\frac{x+{\sec }^2(x)\tan (x)}{\sqrt{x^2+{\sec }^2(x)}}}}\phantom{\frac{dy}{dx}}& \phantom{=\frac{x+{\sec }^2(x)\tan (x)}{\sqrt{x^2+{\sec }^2(x)}}}\end{array}$$

b) $$\begin{array}{rl}y& ={\log }_5(\csc (x))\\ \frac{dy}{dx}& =\frac{1}{\csc (x)\ln (5)}\times \frac{d}{dx}\csc (x)\\ & =\frac{1}{\csc (x)\ln (5)}\times -\cot (x)\csc (x)\\ \boxed{{\displaystyle \frac{dy}{dx}=-\frac{\cot (x)}{\ln (5)}}}\phantom{\frac{dy}{dx}}& \phantom{=-\frac{\cot (x)}{\ln (5)}}\end{array}$$

c) $$\begin{array}{rl}y& ={\tan }^2(\sin (3x))\\ \frac{dy}{dx}& =2\tan (\sin (3x))\times \frac{d}{dx}\tan (\sin (3x))\\ & =2\tan (\sin (3x))\times {\sec }^2(\sin (3x))\times \frac{d}{dx}\sin (3x)\\ & =2\tan (\sin (3x))\times {\sec }^2(\sin (3x))\times \cos (3x)\times \frac{d}{dx}3x\\ & =2\tan (\sin (3x))\times {\sec }^2(\sin (3x))\times \cos (3x)\times 3\\ \boxed{{\displaystyle \frac{dy}{dx}=6\tan (\sin (3x)){\sec }^2(\sin (3x))\cos (3x)}}\phantom{\frac{dy}{dx}}& \phantom{=6\tan (\sin (3x)){\sec }^2(\sin (3x))\cos (3x)}\end{array}$$