# How do you differentiate f(x)=sec^2(e^(x^4) )  using the chain rule?

Mar 14, 2018

${f}^{'} \left(x\right) = 8 {x}^{3} \left({e}^{{x}^{4}}\right) {\sec}^{2} \left({e}^{{x}^{4}}\right) \tan \left({e}^{{x}^{4}}\right)$

#### Explanation:

Chain Rule:
$\textcolor{red}{\frac{d}{\mathrm{dx}} \left(g \circ f \left(x\right)\right) = {g}^{'} \left(f \left(x\right)\right) \cdot {f}^{'} \left(x\right)}$
$f \left(x\right) = {\sec}^{2} \left({e}^{{x}^{4}}\right) = {\left[\textcolor{red}{\sec \left({e}^{{x}^{4}}\right)}\right]}^{2}$
f^'(x)=2sec(e^(x^(4)))*d/(dx)color(red)((seccolor(blue)((e^(x^(4)))))
f^'(x)=2sec(e^(x^(4)))sec(e^(x^(4)))tan(e^(x^(4)))d/(dx)color(blue)((e^(x^4))
${f}^{'} \left(x\right) = 2 \sec \left({e}^{{x}^{4}}\right) \sec \left({e}^{{x}^{4}}\right) \tan \left({e}^{{x}^{4}}\right) \left({e}^{{x}^{4}}\right) \frac{d}{\mathrm{dx}} \left(\textcolor{red}{{x}^{4}}\right)$
${f}^{'} \left(x\right) = 2 \sec \left({e}^{{x}^{4}}\right) \sec \left({e}^{{x}^{4}}\right) \tan \left({e}^{{x}^{4}}\right) \left({e}^{{x}^{4}}\right) \left(4 {x}^{3}\right)$
${f}^{'} \left(x\right) = 8 {x}^{3} \left({e}^{{x}^{4}}\right) \sec \left({e}^{{x}^{4}}\right) \sec \left({e}^{{x}^{4}}\right) \tan \left({e}^{{x}^{4}}\right)$
${f}^{'} \left(x\right) = 8 {x}^{3} \left({e}^{{x}^{4}}\right) {\sec}^{2} \left({e}^{{x}^{4}}\right) \tan \left({e}^{{x}^{4}}\right)$