Differentiation Rules: Product, Quotient and Chain Rule

We start by writing $x=\cos(y)$ and differentiate with respect to $y$: $\frac{dx}{dy}=-\sin(y)$. Hence, it follows that $\frac{dy}{dx}=\frac{1}{-\sin(y)}$. To write this as a function of $x$, we note that $\sin(y)=\pm\sqrt{1-\cos^2(y)}=\pm\sqrt{1-x^2}$. Recall that $\cos(y)$ is one-to-one on the default interval $0\leq y\leq\pi$. It follows that $\sin(y)$ is positive on this interval. Hence $\frac{d}{dx}\left(\arccos(x)\right)=-\frac{1}{\sqrt{1-x^2}}$

As we can see, $f(x)$ is a product that we can write as $f(x)=u(x)v(x)$ where $u(x)=3e^{5x}$ and $v(x)=\cos(2x)$. It follows that $u'(x)=15e^{5x}$ (see differentiating exponential functions) and $v'(x)=-2\sin(2x)$ (see differentiating trigonometric functions). We can then use the product rule:

$\begin{array}{l}f'(x)&=&u'(x)v(x)+u(x)v'(x)\\&=&15e^{5x}\cos(2x)-6e^{5x}\sin(2x)\\&=&3e^{5x}\left(5\cos(2x)-2\sin(2x)\right)\end{array}$

As we can see $y$ is a quotient and we can write $y=\frac{f(x)}{g(x)}$ where $f(x)=2^{3x}$ and $g(x)=\ln(4x)$. It follows that $\frac{df}{dx}=3\times 2^{3x}\ln(2)$ and $\frac{dg}{dx}=\frac{1}{x}$ (see differentiating exp/log functions). It follows, using the quotient rule that

$\begin{array}{l}\frac{dy}{dx}&=&\frac{\frac{df}{dx}g-f\frac{dg}{dx}}{g^2}\\&=&\frac{3\times 2^{3x}\ln(2)\times\ln(4x)-2^{3x}\times\frac{1}{x}}{\ln(4x)^2}\\&=&2^{3x}\left(3\ln(2)\ln(4x)^{-1}-x^{-1}\ln(4x)^{-2}\right)\end{array}$.

Alternatively,  we could write $y=2^{3x}\ln(4x)^{-1}$ and use the product and chain rule. Let $f(x)=2^{3x}$ and $g(x)=\ln(4x)^{-1}$. It follows that $f'(x)=3\times 2^{3x}\ln(2)$ and $g'(x)=-\ln(4x)^{-2}\times x^{-1}$ (via the chain rule). Hence, by the product rule, $\frac{dy}{dx}=3\times 2^{3x}\ln(2)\times \ln(4x)^{-1}+2^{3x}\times-x^{-1}\ln(4x)^{-2}$. This is equivalent to the expression found using the quotient rule.

The function that we must differentiate is a chain. This is because we apply arctan on top of sec of $x$. It follows that we can write $y=\arctan(g)$ where $g(x)=\sec(x)$. We then use the chain rule:

$\begin{array}{l}\frac{dy}{dx}&=&\frac{dy}{dg}\times \frac{dg}{dx}\\&=&\frac{1}{1+g^2}\times \text{sec}(x)\tan(x)\\&=&\frac{1}{1+\sec^2(x)}\times\sec(x)\tan(x)\\&=&\frac{1}{1+\frac{1}{\cos^2(x)}}\times\frac{1}{\cos(x)}\frac{\sin(x)}{\cos(x)}\\&=&\frac{\cos^2(x)}{\cos^2(x)+1}\times\frac{1}{\cos(x)}\frac{\sin(x)}{\cos(x)}\\&=&\frac{\sin(x)}{\cos^2(x)+1}\end{array}$

as required. See more on differentiating trigonometric functions.