Question

Differentiate

$-3x^2.\left(\sin 2x^3\right)\left\{\cos \right[{\cos }^2\left(x^3\right)\left]\right\}$

Answer 1

Let $y=-3x^2\left(\sin 2x^3\right)\cos \left({\cos }^2\right(x^3\left)\right)$

Let $u=-3x^2$

$\therefore \frac{du}{dx}=-6x$

let $v=\sin \left(2x^3\right)$

$\frac{dv}{dx}=\cos \left(2x^3\right)\left(6x^2\right)$

$\frac{dv}{dx}=\left(6x^2\right)\cos \left(2x^3\right)$

Let $w=\cos \left({\cos }^2\right(x^3\left)\right)$

$\frac{dw}{dx}=-\sin \left({\cos }^2\right(x^3\left)\right)\frac{d}{dx}\left({\cos }^2\right(x^3\left)\right)$

$=2\sin \left({\cos }^2\right(x^3\left)\right)\cos \left(x^3\right)\sin \left(x^3\right)\left(3x^2\right)$

$\frac{dw}{dx}=\left(3x^2\right)\sin \left(2x^3\right)\sin \left({\cos }^2\right(x^3\left)\right)$

$y=uvw$

Differentiate on both sides w.r.t x

$\frac{dy}{dx}=\frac{du}{dx}\times v\times w+u\times \frac{dv}{dx}\times w+u\times v\times \frac{dw}{dx}$

$\frac{dy}{dx}=(-6x)\left(\sin \right(2x^3\left)\right)\cos \left[{\cos }^2\right(x^3\left)\right]+(-3x^2)\left(6x^2\right)\cos \left(2x^3\right)\cos \left[{\cos }^2\right(x^3\left)\right]+(-3x^2)\left(\sin \right(2x^3\left)\right)\left(3x^2\right)\left(\sin \right(2x^3\left)\right)\times$

$\sin \left[{\cos }^2\right(x^3\left)\right]$

$\therefore \frac{dy}{dx}=(-3x^2)\left[2{\sin }^2\right(x^3)\cos \left[{\cos }^2\right(x^3\left)\right]+6x^3\cos (2x^3)\cos \left[{\cos }^2\right(x^3\left)\right]+3x^3{\sin }^2(2x^3\left)\sin \left[{\cos }^2\right(x^3\left)\right]\right]$