MCQ
If $y = {3^{{x^2}}}$, then ${{dy} \over {dx}}$ is equal to
- A$({x^2}){3^{{x^2} - 1}}$
- B$3{x^2}.2x$
- ✓${3^{{x^2}}}.2x.\log 3$
- D$({x^2} - 1).3$
$\because \frac{d}{{dx}}({a^x}) = {a^x}{\log _e}a$
$\therefore \frac{{dy}}{{dx}} = {3^{{x^2}}}{\log _e}3\frac{d}{{dx}}({x^2})$
==> $\frac{{dy}}{{dx}} = {3^{{x^2}}}.2x.{\log _e}3$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$( S_{1})$: $2|\hat{ a } \times \hat{ b }|=|\hat{ a }-\hat{ b }|$
$(S_{2})$ : The projection of $\hat{a}$ on $(\hat{a}+\hat{b})$ is $\frac{1}{2}$