How do you differentiate f(x)=sin(1/sqrt(3x^2-4) )  using the chain rule?

Jan 21, 2016

$f ' \left(x\right) = - \cos \left(\frac{1}{3 {x}^{2} - 4}\right) \cdot \frac{6 x}{2 \sqrt{{\left(3 {x}^{2} - 4\right)}^{3}}}$

Explanation:

Let's break down your function using the chain rule:

$f \left(x\right) = \sin \left(\textcolor{b l u e}{\frac{1}{\sqrt{3 {x}^{2} - 4}}}\right) = \sin \textcolor{b l u e}{u}$

where $\text{ } u = \frac{1}{\sqrt{\textcolor{\mathmr{and} a n \ge}{3 {x}^{2} - 4}}} = \frac{1}{\sqrt{\textcolor{\mathmr{and} a n \ge}{v}}}$

where $\text{ } v = 3 {x}^{2} - 4$

According to the chain rule, the derivative is:

$f ' \left(x\right) = \left[\sin u\right] ' \cdot u ' = \left[\sin u\right] ' \cdot \left[\frac{1}{\sqrt{v}}\right] ' \cdot v '$

Let's compute the derivatives of those terms!

$\left[\sin u\right] ' = \cos u = \cos \left(\frac{1}{3 {x}^{2} - 4}\right)$

$u ' = \left[\frac{1}{\sqrt{v}}\right] ' = \left[{v}^{- \frac{1}{2}}\right] ' = - \frac{1}{2} {v}^{- \frac{3}{2}} = - \frac{1}{2 \sqrt{{v}^{3}}} = - \frac{1}{2 \sqrt{{\left(3 {x}^{2} - 4\right)}^{3}}}$

$v ' = \left[3 {x}^{2} - 4\right] ' = 6 x$

Thus, your derivative is:

$f ' \left(x\right) = \left[\sin u\right] ' \cdot \left[\frac{1}{\sqrt{v}}\right] ' \cdot v '$

$= \cos \left(\frac{1}{3 {x}^{2} - 4}\right) \cdot \left(- \frac{1}{2 \sqrt{{\left(3 {x}^{2} - 4\right)}^{3}}}\right) \cdot 6 x$

$= - \cos \left(\frac{1}{3 {x}^{2} - 4}\right) \cdot \frac{6 x}{2 \sqrt{{\left(3 {x}^{2} - 4\right)}^{3}}}$