Question

Differentiate the following functions with respect to x:
(i) ln(1 + sin2 x) (ii) x²

Answer 1

(i)

Let y = ln(1+sin2x)

Also let z = 1+sin2x

So y = ln(z)

Differentiating with respect to z

$\frac{dy}{dz}$ = $\frac{1}{z}$

Again z = 1+sin2x

So $\frac{dz}{dx}$ = 2cos2x as $\frac{d(1)}{dx} = 0$ and $, \frac{d}{dx}[sin(mx)] = mcos(mx)$

Now $\frac{dy}{dx} = \frac{dy}{dz} \frac{dz}{dx}$ = $\frac{1}{z}$ (2cos 2x)

=> $\frac{dy}{dx} = \frac{2cos2x}{1+sin2x}$

You can further simplify as follows

$\frac{dy}{dx} = \frac{2cos2x(1-sin2x)}{1-sin^22x}$

=> $\frac{dy}{dx} = \frac{2cos2x(1-sin2x)}{cos^22x}$

=> $\frac{dy}{dx} = \frac{2(1-sin2x)}{cos2x}$

=> $\frac{dy}{dx} = 2(sec2x-tan2x)$

Therefore

$\frac{d}{dx}(ln(1+sin2x)$

= $2(sec2x-tan2x)$

(ii)

Let y = x²

We know that $\frac{d}{dx}(x^n) = nx^{n-1}$

So $\frac{dy}{dx} = 2x^{2-1}$

=> $\frac{dy}{dx} = 2x$

So $\frac{d}{dx}(x^2) = 2x$