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) \[ \begin{align} y &= \sqrt{x^2 + \sec^2(x)} = \left(x^2 + \sec^2(x)\right)^{1/2} \\ \frac{dy}{dx} &= \frac{1}{2}\left(x^2 + \sec^2(x)\right)^{-1/2} \times \frac{d}{dx}\left(x^2 + \sec^2(x)\right) \\ &= \frac{1}{2}\left(x^2 + \sec^2(x)\right)^{-1/2} \times \left(2x + 2\sec(x)\times \frac{d}{dx} \sec(x)\right) \\ &= \frac{1}{2}\left(x^2 + \sec^2(x)\right)^{-1/2} \times \left(2x + 2\sec^2(x)\tan(x)\right) \\ \Aboxed{\frac{dy}{dx}&= \frac{x + \sec^2(x)\tan(x)}{\sqrt{x^2 + \sec^2(x)}}} \end{align} \]

b) \[ \begin{align} y &= \log_5(\csc(x)) \\ \frac{dy}{dx} &= \frac{1}{\csc(x)\ln(5)} \times \frac{d}{dx} \csc(x) \\ &= \frac{1}{\csc(x)\ln(5)} \times -\cot(x)\csc(x) \\ \Aboxed{\frac{dy}{dx} &= -\frac{\cot(x)}{\ln(5)}} \end{align} \]

c) \[ \begin{align} y &= \tan^2(\sin(3x)) \\ \frac{dy}{dx} &= 2\tan(\sin(3x)) \times \frac{d}{dx} \tan(\sin(3x)) \\ &= 2\tan(\sin(3x)) \times \sec^2(\sin(3x)) \times \frac{d}{dx} \sin(3x) \\ &= 2\tan(\sin(3x)) \times \sec^2(\sin(3x)) \times \cos(3x) \times \frac{d}{dx} 3x \\ &= 2\tan(\sin(3x)) \times \sec^2(\sin(3x)) \times \cos(3x) \times 3 \\ \Aboxed{\frac{dy}{dx} &= 6\tan(\sin(3x))\sec^2(\sin(3x))\cos(3x)} \end{align} \]