Answer
a) $\frac{2\csc(2x)\cot(2x)(x^{3}+5)+3x^{2}\csc(2x)}{(x^{3}+5)^2}\times \csc^{2}\left(\frac{\csc(2x)}{x^{3}+5}\right)$
b) $-\frac{2+3\sin^{2}(x)\cos(x)}{(2x+\sin^{3}(x))^2}$
Work Step by Step
a) We have: $\cot\left(\frac{\csc(2x)}{x^{3}+5}\right)=f(x)$
$\implies$ First, we will differentiate the argument for cot $w.r.t.x$
Let the argument for cot be $A$.
$\implies \frac{dA}{dx}=\frac{d\frac{\csc(2x)}{x^{3}+5}}{dx}= \frac{-2\csc(2x)\cot(2x)(x^{3}+5)-3x^{2}\csc(2x)}{(x^{3}+5)^2}\quad$(Quotient Rule)
$\implies \frac{d \cot(A)}{dx} = \frac{d \cot(A)}{dA}\times\frac{dA}{dx}$ (Chain Rule)
$\quad = \frac{2\csc(2x)\cot(2x)(x^{3}+5)+3x^{2}\csc(2x)}{(x^{3}+5)^2}\times \csc^{2}(\frac{\csc(2x)}{x^{3}+5})$
b) We have: $f(x) = \frac{1}{2x+\sin^{3}(x)}$
Differentiating $f(x)\space w.r.t.x:$
Let $2x+\sin^{3}(x) = A, \sin^{3}(x) =B$.
$\implies f'(x) = \frac{d(\frac{1}{A})}{dA}\times\frac{dA}{dx} = -\frac{1}{(2x+\sin^{3}(x))^2}\times\frac{dA}{dx}\rightarrow ①$
$\implies \frac{dA}{dx} = \frac{d(2x+B)}{dx}=2+\frac{dB}{dx}\rightarrow②$
$\implies \frac{dB}{dx} = \frac{d(\sin^{3}(x))}{d(\sin(x))}\times\frac{d(\sin(x)}{dx} = 3\sin^{2}(x)\times \cos(x)$
Substituting this in $②$:
$\implies \frac{dA}{dx} = 2+3\sin^{2}(x)\times \cos(x)$
Substituting this in $①$:
$\therefore f'(x) = -\frac{1}{(2x+\sin^{3}(x))^2}\times2+3\sin^{2}(x)\times \cos(x)$
