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

Jul 4, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos {x}^{3} \left[\cos {x}^{3} - 6 {x}^{3} \sin {x}^{3}\right]$

Explanation:

Here,

$y = \frac{x}{\sec} ^ 2 \left({x}^{3}\right)$

$\implies y = x \cdot {\cos}^{2} \left({x}^{3}\right)$

Diff.w.r.t. $x$ , using Product rule and Chain Rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \cdot \frac{d}{\mathrm{dx}} \left({\cos}^{2} {x}^{3}\right) + {\cos}^{2} {x}^{3} \cdot \frac{d}{\mathrm{dx}} \left(x\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = x \cdot 2 \cos {x}^{3} \left(- \sin {x}^{3}\right) \frac{d}{\mathrm{dx}} \left({x}^{3}\right) + {\cos}^{2} {x}^{3} \cdot 1$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x \sin {x}^{3} \cos {x}^{3} \cdot 3 {x}^{2} + {\cos}^{2} {x}^{3}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 6 {x}^{3} \sin {x}^{3} \cos {x}^{3} + {\cos}^{2} {x}^{3}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} {x}^{3} - 6 {x}^{3} \sin {x}^{3} \cos {x}^{3.} . . \to \left(1\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \cos {x}^{3} \left[\cos {x}^{3} - 6 {x}^{3} \sin {x}^{3}\right]$

Note :

From $\left(1\right)$ we can simplify in the form :

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} {x}^{3} - 6 {x}^{3} \sin {x}^{3} \cos {x}^{3}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} {x}^{3} - 3 {x}^{3} \left\{2 \sin {x}^{3} \cos {x}^{3}\right\}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} {x}^{3} - 3 {x}^{3} \cdot \sin 2 \left({x}^{3}\right)$