# How do you differentiate f(x)=tan(sqrt(x^3-1))  using the chain rule?

Jul 20, 2017

f'(x)= (3x^2sec^2(sqrt(x^3-1)))/(2sqrt(x^3-1)

#### Explanation:

$f \left(x\right) = \tan \left(\sqrt{{x}^{3} - 1}\right)$

To find $f '$, set $u = \sqrt{{x}^{3} - 1}$ and $u ' = \frac{3 {x}^{2}}{2 \sqrt{{x}^{3} - 1}}$ and $f ' \left(u\right) = {\sec}^{2} u = {\sec}^{2} \left(\sqrt{{x}^{3} - 1}\right)$so that f'(x) = f'(u) xx u'=(3x^2sec^2(sqrt(x^3-1)))/(2sqrt(x^3-1)

Jul 20, 2017

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

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given " y=f(g(h(x)))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(h \left(x\right)\right)\right) . g ' \left(h \left(x\right)\right) . h ' \left(x\right) \leftarrow \text{ chain rule}$

$y = \tan \left(\sqrt{{x}^{3} - 1}\right)$

$\Rightarrow f ' \left(g \left(h \left(x\right)\right)\right) = {\sec}^{2} \left(\sqrt{{x}^{3} - 1}\right) \to \left(\textcolor{red}{1}\right)$

$g \left(h \left(x\right)\right) = {\left({x}^{3} - 1\right)}^{\frac{1}{2}} \Rightarrow g ' \left(h \left(x\right)\right) = \frac{1}{2} {\left({x}^{3} - 1\right)}^{- \frac{1}{2}} \to \left(\textcolor{red}{2}\right)$

$h \left(x\right) = {x}^{3} - 1 \Rightarrow h ' \left(x\right) = 3 {x}^{2} \to \left(\textcolor{red}{3}\right)$

$\text{combining the product of all 3 parts gives}$

$\Rightarrow f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} {\sec}^{2} \left(\sqrt{{x}^{3} - 1}\right)}{2 \sqrt{{x}^{3} - 1}}$