# How do you differentiate f(x)=1/sin(sqrt(1/x)) using the chain rule?

Dec 30, 2015

$f ' \left(x\right) = \frac{\cos \left(\frac{1}{\sqrt{x}}\right)}{2 {\sin}^{2} \left(\frac{1}{\sqrt{x}}\right) {x}^{\frac{3}{2}}}$

#### Explanation:

$f \left(x\right) = {\left(\sin \left({x}^{- \frac{1}{2}}\right)\right)}^{-} 1$

Use the chain rule: $\frac{d}{\mathrm{dx}} {u}^{-} 1 = - {u}^{-} 2 \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

f'(x)=-sin^-2(x^(-1/2))*color(blue)(d/dxsin(x^(-1/2))

To find this derivative, use chain rule again:

$\frac{d}{\mathrm{dx}} \sin \left(u\right) = \cos \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Thus,

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \sin \left({x}^{- \frac{1}{2}}\right)} = \cos \left({x}^{- \frac{1}{2}}\right) \cdot \frac{d}{\mathrm{dx}} {x}^{- \frac{1}{2}}$

=>color(blue)(-1/2x^(-3/2)cos(x^(-1/2))

Plug this back in.

$f ' \left(x\right) = - {\sin}^{-} 2 \left({x}^{- \frac{1}{2}}\right) \cdot - \frac{1}{2} {x}^{- \frac{3}{2}} \cos \left({x}^{- \frac{1}{2}}\right)$

$\implies \frac{\cos \left(\frac{1}{\sqrt{x}}\right)}{2 {\sin}^{2} \left(\frac{1}{\sqrt{x}}\right) {x}^{\frac{3}{2}}}$