# How do you differentiate f(x)=sec^4(e^(x^3) )  using the chain rule?

Jan 28, 2016

$f ' \left(x\right) = 12 {x}^{2} {e}^{{x}^{3}} {\sec}^{4} \left({e}^{{x}^{3}}\right) \tan \left({e}^{{x}^{3}}\right)$

#### Explanation:

The first issue is the fourth power. We can deal with it through the application of the chain rule $\frac{d}{\mathrm{dx}} \left({u}^{4}\right) = 4 {u}^{3} \cdot u '$, where $u = \sec \left({e}^{{x}^{3}}\right)$, yielding

$f ' \left(x\right) = 4 {\sec}^{3} \left({e}^{{x}^{3}}\right) \cdot \frac{d}{\mathrm{dx}} \left(\sec \left({e}^{{x}^{3}}\right)\right)$

To differentiate the secant function, use the rule: $\frac{d}{\mathrm{dx}} \left(\sec \left(u\right)\right) = \sec \left(u\right) \tan \left(u\right) \cdot u '$, where now $u = {e}^{{x}^{3}}$. This gives

$f ' \left(x\right) = 4 {\sec}^{3} \left({e}^{{x}^{3}}\right) \sec \left({e}^{{x}^{3}}\right) \tan \left({e}^{{x}^{3}}\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{{x}^{3}}\right)$

Which simplifies to be

$f ' \left(x\right) = 4 {\sec}^{4} \left({e}^{{x}^{3}}\right) \tan \left({e}^{{x}^{3}}\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{{x}^{3}}\right)$

All that remains is the differentiation of ${e}^{{x}^{3}}$, to which the chain rule will again be applied: $\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot u '$. Now, $u = {x}^{3}$, so

$f ' \left(x\right) = 4 {\sec}^{4} \left({e}^{{x}^{3}}\right) \tan \left({e}^{{x}^{3}}\right) \cdot {e}^{{x}^{3}} \frac{d}{\mathrm{dx}} \left({x}^{3}\right)$

Since $\frac{d}{\mathrm{dx}} \left({x}^{3}\right) = 3 {x}^{2}$,

$f ' \left(x\right) = 12 {x}^{2} {e}^{{x}^{3}} {\sec}^{4} \left({e}^{{x}^{3}}\right) \tan \left({e}^{{x}^{3}}\right)$