Quiz 5 - More chain rule and differentiation of trigonometric functions

Problem 1

Question

Differentiate \(y = \sqrt{x}^x\).

Solution

\[ \begin{align} y &= \sqrt{x}^x \\ \ln(y) &= \ln(\sqrt{x}^x) \\ \frac{d}{dx}\ln(y) &= \frac{d}{dx}\ln(\sqrt{x}^x) \\ \frac{y'}{y} &= \frac{d}{dx}\left(x\ln(\sqrt{x})\right) \\ y' &= y\left(\ln(\sqrt{x}) + \frac{x}{\sqrt{x}}\frac{d}{dx}\sqrt{x}\right) \\ &= \sqrt{x}^x\left(\ln(\sqrt{x}) + \frac{x}{\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)\right) \\ &= \sqrt{x}^x\left(\frac{1}{2}\ln(x) + \frac{1}{2}\right) \\ \Aboxed{y' &= \frac{1}{2}\sqrt{x}^x\left(\ln(x) + 1\right)} \end{align} \]

Problem 2

Question

Differentiate \(y = 10^{x + \cos(x)}\).

Solution

\[ \begin{align} y &= 10^{x + \cos(x)} \\ \ln(y) &= (x + \cos(x))\ln(10) \\ \frac{d}{dx}\ln(y) &= \ln(10)\frac{d}{dx}(x + \cos(x)) \\ \frac{y'}{y} &= \ln(10)(1 - \sin(x)) \\ y' &= y\ln(10)(1 - \sin(x)) \\ \Aboxed{y' &= 10^{x+\cos(x)}\ln(10)(1 - \sin(x))} \end{align} \]

Problem 3

Question

Differentiate \(y = \frac{\ln(x)}{x^e + e^e}\).

Solution

\[ \begin{align} y &= \frac{\ln(x)}{x^e + e^e} \\ y' &= \frac{\frac{1}{x}(x^e + e^e) - \ln(x)(ex^{e-1})}{(x^e+e^e)^2}\\ &= \frac{\frac{1}{x}(x^e + e^e) - \ln(x)(ex^{e-1})}{(x^e+e^e)^2}\\ \Aboxed{y' &= \frac{x^e + e^e - \ln(x)(ex^e)}{x(x^e+e^e)^2}} \end{align} \]