How do you differentiate f(x)=csc(sqrt(2x))  using the chain rule?

Jan 2, 2016

$f ' \left(x\right) = \frac{- \csc \left(\sqrt{2 x}\right) \cot \left(\sqrt{2 x}\right)}{\sqrt{2 x}}$

Explanation:

According to the chain rule,

$\frac{d}{\mathrm{dx}} \left(\csc u\right) = - u ' \csc u \cot u$

Thus,

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\csc \left(\sqrt{2 x}\right)\right) = - \csc \left(\sqrt{2 x}\right) \cot \left(\sqrt{2 x}\right) \cdot \frac{d}{\mathrm{dx}} \left(\sqrt{2 x}\right)$

To find $\frac{d}{\mathrm{dx}} \left(\sqrt{2 x}\right)$, use the chain rule again:

$\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2 \sqrt{u}} \cdot u '$

So,

d/dx(sqrt(2x))=1/(2sqrt(2x))*2=1/(sqrt(2x)

Plug this back in to find $f ' \left(x\right)$:

$f ' \left(x\right) = \frac{- \csc \left(\sqrt{2 x}\right) \cot \left(\sqrt{2 x}\right)}{\sqrt{2 x}}$