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

##### 1 Answer
Nov 14, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \left({x}^{2} \cdot \tan \left({x}^{2}\right) + \ln \left(\sec \left({x}^{2}\right)\right)\right)$

#### Explanation:

So we have

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

Derivating and using the product rule we have

dy/dx = x^2d/dx(ln(sec(x^2)) + ln(sec(x^2))d/dxx^2

dy/dx = x^2d/dx(ln(sec(x^2)) + 2xln(sec(x^2))

Let's say $\sec \left({x}^{2}\right) = u$ then, by the chain rule we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \frac{d}{\mathrm{dx}} \left(\ln \left(u\right)\right) + 2 x \ln \left(\sec \left({x}^{2}\right)\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \frac{d}{\mathrm{du}} \left(\ln \left(u\right)\right) \frac{\mathrm{du}}{\mathrm{dx}} + 2 x \ln \left(\sec \left({x}^{2}\right)\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} / u \cdot \frac{d}{\mathrm{dx}} \left(\sec \left({x}^{2}\right)\right) + 2 x \ln \left(\sec \left({x}^{2}\right)\right)$

Let's say ${x}^{2} = v$ then we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} / u \cdot \frac{d}{\mathrm{dv}} \sec \left(v\right) \cdot \frac{\mathrm{dv}}{\mathrm{dx}} + 2 x \ln \left(\sec \left({x}^{2}\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} / u \cdot \tan \left(v\right) \cdot \sec \left(v\right) \cdot 2 x + 2 x \ln \left(\sec \left({x}^{2}\right)\right)$

Putting it all in terms of $x$ we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \left({x}^{2} / \sec \left({x}^{2}\right) \cdot \tan \left({x}^{2}\right) \cdot \sec \left({x}^{2}\right) + \ln \left(\sec \left({x}^{2}\right)\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \left({x}^{2} \cdot \tan \left({x}^{2}\right) + \ln \left(\sec \left({x}^{2}\right)\right)\right)$