Question

Differentiate the following with regard to x

y=ln( square toot X multiple e^x^2)

Answer 1

$y=ln( \sqrt {x} \times e^{x^2})$

Let $u=\sqrt {x} \times e^{x^2}\implies y=ln(u)\implies {dy\over du}={1\over u}$

Let $f=\sqrt x=x^{1\over 2} \implies f'={1\over 2}x^{{1\over 2}-1}={1\over 2}x^{-{1\over 2}}$

Let $g=e^{x^2} \implies g'=2xe^{x^2}$

We then use product rule of differentiation to differentiate $u=\sqrt {x} \times e^{x^2}$ with respect to $x$

${du \over dx}=f'g+fg'$

$={1\over 2}x^{-{1\over 2}}e^{x^2}+\sqrt x(2xe^{x^2})$

We then use chain rule of differentiation to differentiate $y=ln( \sqrt {x} \times e^{x^2})$ with respect to $x$

${dy\over dx}={dy\over du}\times{du\over dx}$

$={1\over u}\times({1\over 2}x^{-{1\over 2}}e^{x^2}+\sqrt x(2xe^{x^2}))$

$={1\over \sqrt {x} \times e^{x^2}}\times({1\over 2}x^{-{1\over 2}}e^{x^2}+\sqrt x(2xe^{x^2}))$

$={1\over \sqrt {x} \times e^{x^2}}\times({1\over 2(\sqrt x)}e^{x^2}+\sqrt x(2xe^{x^2}))$

$={e^{x^2}\over 2xe^{x^2}}+{\sqrt x(2xe^{x^2}\over \sqrt x(e^{x^2})}$

$={1\over 2x}+2x$

$\therefore {dy\over dx}={1+4x^2\over 2x}$