Question

Differentiate w.r.t x:

$y=x^2\ln \left(\sqrt{\frac{x^2+9}{x^2+4}}\right)$

Answer 1

Let $y=x^2\ln \left(\sqrt{\frac{x^2+9}{x^2+4}}\right)$

Let $t=\ln \left(\sqrt{\frac{x^2+9}{x^2+4}}\right)$

$e^t=\sqrt{\frac{x^2+9}{x^2+4}}$

$e^{2t}=\frac{x^2+9}{x^2+4}$

Differentiate on both sides w.r.t. x

$e^{2t}.2.\frac{dt}{dx}=\frac{2x(x^2+4)-2x(x^2+9)}{{(x^2+4)}^2}$

$\frac{(x^2+9)}{(x^2+4)}.2.\frac{dt}{dx}=\frac{-10x}{{(x^2+4)}^2}$

$\therefore \frac{dt}{dx}=\frac{-5x}{(x^2+4)(x^2+9)}$

$y=x^{2}t$

Differentiate on both sides w.r.t. x

$\frac{dy}{dx}=2xt+x^2\frac{dt}{dx}$

$\frac{dy}{dx}=2x\ln \left(\sqrt{\frac{x^2+9}{x^2+4}}\right)-\frac{5x^3}{(x^2+4)(x^2+9)}$