# If f(x) =csc^3(x/2)  and g(x) = sqrt(2x+3 , what is f'(g(x)) ?

Apr 28, 2017

$\textcolor{red}{f ' \left(g \left(x\right)\right) = - \frac{3}{2 \sqrt{2 x + 3}} \cdot {\csc}^{3} \left(\frac{\sqrt{2 x + 3}}{2}\right) \cdot \cot \left(\frac{\sqrt{2 x + 3}}{2}\right)}$

#### Explanation:

$f \left(x\right) = {\csc}^{3} \left(\frac{x}{2}\right)$

$g \left(x\right) = \sqrt{2 x + 3}$

$\therefore f \left(g \left(x\right)\right) = {\csc}^{3} \left(\frac{\sqrt{2 x + 3}}{2}\right)$

$\implies f ' \left(x\right) = \frac{d \left[{\csc}^{3} \left(\frac{\sqrt{2 x + 3}}{2}\right)\right]}{\mathrm{dx}}$

Using chain rule to evaluate $\frac{d \left[{\csc}^{3} \left(\frac{\sqrt{2 x + 3}}{2}\right)\right]}{\mathrm{dx}}$

$\implies f ' \left(x\right) = \frac{d \left[{\csc}^{3} \left(\frac{\sqrt{2 x + 3}}{2}\right)\right]}{d \left[\csc \left(\frac{\sqrt{2 x + 3}}{2}\right)\right]} \times \frac{d \left[\csc \left(\frac{\sqrt{2 x + 3}}{2}\right)\right]}{d \left(\frac{\sqrt{2 x + 3}}{2}\right)} \times \frac{d \left(\frac{\sqrt{2 x + 3}}{2}\right)}{d \left(2 x + 3\right)} \times \frac{d \left(2 x + 3\right)}{\mathrm{dx}}$

Now,

1.

[d[csc^3(sqrt(2x+3)/2)]]/(d[csc(sqrt(2x+3)/2)]) = 3*csc^(3-1)(sqrt(2x+3)/2) =color(green){3csc^2(sqrt(2x+3)/2)

2.

$\frac{d \left[\csc \left(\frac{\sqrt{2 x + 3}}{2}\right)\right]}{d \left(\frac{\sqrt{2 x + 3}}{2}\right)} = \textcolor{g r e e n}{- \csc \left(\frac{\sqrt{2 x + 3}}{2}\right) \cdot \cot \left(\frac{\sqrt{2 x + 3}}{2}\right)}$

3.

$\frac{d \left(\frac{\sqrt{2 x + 3}}{2}\right)}{d \left(2 x + 3\right)} = \frac{1}{2} \frac{d \left(\sqrt{2 x + 3}\right)}{d \left(2 x + 3\right)} = \frac{1}{2} \cdot \frac{d {\left(2 x + 3\right)}^{\frac{1}{2}}}{d \left(2 x + 3\right)}$

$= \frac{1}{2} \cdot \frac{1}{2} \cdot {\left(2 x + 3\right)}^{\frac{1}{2} - 1} = \frac{1}{4} \cdot {\left(2 x + 3\right)}^{- \frac{1}{2}} = \textcolor{g r e e n}{\frac{1}{4 \sqrt{2 x + 3}}}$

4.

$\frac{d \left(2 x + 3\right)}{\mathrm{dx}} = \frac{d \left(2 x\right)}{\mathrm{dx}} + \frac{d \left(3\right)}{\mathrm{dx}}$ . [Using sum rule]
$= 2 \frac{\mathrm{dx}}{\mathrm{dx}} + 0 = \textcolor{g r e e n}{2}$

therefore f'(g(x))=color(green){3csc^2(sqrt(2x+3)/2) xx color(green){-csc(sqrt(2x+3)/2)*cot(sqrt(2x+3)/2)} xx color(green)[1/(4sqrt(2x+3))] xx color(green)2

$= \textcolor{red}{- \frac{3}{2 \sqrt{2 x + 3}} \cdot {\csc}^{3} \left(\frac{\sqrt{2 x + 3}}{2}\right) \cdot \cot \left(\frac{\sqrt{2 x + 3}}{2}\right)}$

Hence,
$\textcolor{red}{f ' \left(g \left(x\right)\right) = - \frac{3}{2 \sqrt{2 x + 3}} \cdot {\csc}^{3} \left(\frac{\sqrt{2 x + 3}}{2}\right) \cdot \cot \left(\frac{\sqrt{2 x + 3}}{2}\right)}$