# How do you differentiate  y =sqrt(sec ^2x^3 using the chain rule?

Mar 5, 2017

$= \frac{3 {x}^{2} \sin {x}^{3} \sqrt{{\cos}^{2} {x}^{3}}}{\cos} ^ 3 {x}^{3}$

#### Explanation:

Let's apply $\sec x = \frac{1}{\cos} x$ and then differentiate:

y=sqrt(1/cos^2x^3

Since

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

then

$y ' = \frac{1}{\cancel{2} \sqrt{\frac{1}{\cos} ^ 2 {x}^{3}}} \cdot \left(- \frac{\cancel{2}}{\cos} ^ 3 {x}^{3}\right) \cdot \left(- \sin {x}^{3}\right) \cdot 3 {x}^{2}$

$= \frac{3 {x}^{2} \sin {x}^{3} \sqrt{{\cos}^{2} {x}^{3}}}{\cos} ^ 3 {x}^{3}$