How do I differentiate this?

Question

anonymous · Answer

$$\frac{ 1 }{ f \sqrt[]{x} }$$

solomonzelman · Answer

Wait, maybe it is:

$\large\color{black}{ \displaystyle \frac{ 1}{\sqrt{x} }  }$   ?

anonymous · Answer

The question says what is the derivative of $$g(x)=\frac{ 1 }{ f(\sqrt{x}) }$$

anonymous · Answer

suppose $f(x)=\sin(x)$ making $f(\sqrt{x})=\sin(\sqrt{x})$ 
how would you take the derivative of $$\frac{1}{\sin(\sqrt{x})}$$?

solomonzelman · Answer

$\large\color{black}{ \displaystyle g(x)=\left\{~f(h(x))~\right\}^n }$   (when n is not 0)

First, derivative of   $x^n$   is   $nx^{n-1}$.

$\large\color{black}{ \displaystyle n\left\{~f(h(x))~\right\}^{n-1}  }$

then you apply the chain rule, knowing that the derivative of f(x) is f'(x) you get:

$\large\color{black}{ \displaystyle n\left\{~f(h(x))~\right\}^{n-1} \times f'(h(x)) }$

and then once again you apply the chain rule knowing that the derivative of h(x) is h'(x).

$\large\color{black}{ \displaystyle g'(x)= n\left\{~f(h(x))~\right\}^{n-1} \times f'(h(x)) \times h'(x)  }$

phi · Answer

you might want to first write the expression as
$$ g(x) = \left( f\left(x^\frac{1}{2}\right)\right)^{-1} $$
now use Solomon's ideas