Question

Differentiate the function w.r.t.x:
$\rm \frac{1}{\sin x}+2^{x+3}+\frac{1}{\log_{x}3 }$

Answer 1

we know that 
$\begin{lgathered}\frac{1}{sin \: x} = cosec \: x \\ \\ log_{x}(3) = \frac{log \: 3}{log \: x} \\ \\\end{lgathered}$
So for derivatives

$\begin{lgathered}\frac{d(cosec \: x)}{dx} = -cosec \: x \: cot \: x \\ \\ \frac{d( {2}^{(x + 3)} )}{dx} = {2}^{(x + 3)} log \: 2 \: \: (1) \\ \\ \frac{d}{dx} ( \frac{log \: x}{log \: 3} ) \\ =\frac{1}{log 3} \: \frac{d}{dx} (log \: x) = \frac{1}{ x\:log 3} \\ \\\end{lgathered}$

put all these derivatives in one expression for final answer

$\begin{lgathered}\frac{d}{dx} \bigg(\frac{1}{sin x}+2^{x+3}-\frac{1}{\log_{x} 3}\bigg) \\\\= -cosec \: x \: cot\: x + {2}^{(x + 3)} log \: 2- \frac{1}{ x\:log 3} \\ \\\end{lgathered}$

Hope it helps you.