MCQ
જો $y = {3^{{x^2}}}$, તો ${{dy} \over {dx}} = . . . .$
- A$({x^2}){3^{{x^2} - 1}}$
- B$3{x^2}.2x$
- ✓${3^{{x^2}}}.2x.\log 3$
- D$({x^2} - 1).3$
$\because \frac{d}{{dx}}({a^x}) = {a^x}{\log _e}a$
$\therefore \frac{{dy}}{{dx}} = {3^{{x^2}}}{\log _e}3\frac{d}{{dx}}({x^2})$
==> $\frac{{dy}}{{dx}} = {3^{{x^2}}}.2x.{\log _e}3$.
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$f(x)=\left\{\begin{array}{ll}-55 x, & \text { if } x<-5 \\ 2 x^{3}-3 x^{2}-120 x, & \text { if }-5 \leq x \leq 4 \\ 2 x^{3}-3 x^{2}-36 x-336, & \text { if } x>4\end{array}\right.$
જો $A=\{ x \in R : f$ એ વધતુ વિધેય છે $\},$ તો $A = ......$