# How do you differentiate f(x)=sqrtsin(1/lnx^2) using the chain rule?

Nov 17, 2015

So, let's break this down into a chain of functions:

$f \left(u\right) = \sqrt{u}$

$u \left(v\right) = \sin \left(v\right)$

$v \left(w\right) = \frac{1}{w}$

$w \left(z\right) = \ln z$

$z \left(x\right) = {x}^{2}$

Now, we need to differentiate every one of those functions:

 f(u) = sqrt(u) = u^(1/2) color(white)(xxx) => color(white)(xx) f'(u) = 1/2 u ^(-1/2) = 1/ (2 sqrt(u)

$u \left(v\right) = \sin \left(v\right) \textcolor{w h i t e}{\times \times \xi i} \implies \textcolor{w h i t e}{\times} u ' \left(v\right) = \cos \left(v\right)$

$v \left(w\right) = \frac{1}{w} \textcolor{w h i t e}{\times \times \times \xi i} \implies \textcolor{w h i t e}{\times} v ' \left(w\right) = - \frac{1}{w} ^ 2$

$w \left(z\right) = \ln z \textcolor{w h i t e}{\times \times \times i} \implies \textcolor{w h i t e}{\times} w ' \left(z\right) = \frac{1}{z}$

$z \left(x\right) = {x}^{2} \textcolor{w h i t e}{\times \times \times \times} \implies \textcolor{w h i t e}{\times} z ' \left(x\right) = 2 x$

So, the derivative of $f \left(x\right)$ is the product of all that derivatives:

$f ' \left(x\right) = f ' \left(u\right) \cdot u ' \left(v\right) \cdot v ' \left(w\right) \cdot w ' \left(z\right) \cdot z ' \left(x\right)$

$\textcolor{w h i t e}{\times \xi i i} = \frac{1}{2 \sqrt{\sin \left(\frac{1}{\ln {x}^{2}}\right)}} \cdot \cos \left(\frac{1}{\ln {x}^{2}}\right) \cdot \left(- \frac{1}{\ln} ^ 2 {x}^{2}\right) \cdot \frac{1}{x} ^ 2 \cdot 2 x$

$\textcolor{w h i t e}{\times \xi i i} = - \frac{\cos \left(\frac{1}{\ln {x}^{2}}\right)}{\sqrt{\sin \left(\frac{1}{\ln {x}^{2}}\right)} \cdot {\ln}^{2} {x}^{2} \cdot x}$