MCQ
If $y ={x^{{x^2}}}$ then $\frac{{dy}}{{dx}}=$
- A$2 \, ln \, x .{x^{{x^2}}}$
- B$(2 \, ln \, x + 1).{x^{{x^2}}}$
- C$(2 \, ln \, x + 1).{x^{{x^2+2}}}$
- ✓${x^{{x^2}}}. \, ln \, ex^2$
Taking log on both the sides, we get $\log y=x \log x$
On differentiating w.r.t. $x$, we get $\frac{1}{y} \frac{d y}{d x}=\frac{x}{x}+\log x$
$\Rightarrow \frac{d y}{d x}=y+y \log x$
$\Rightarrow \frac{d y}{d x}=x^{x}(1+\log x) \ldots\left(\because y=x^{x}\right)$
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$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{x \cos x}\right)\left(\sin ^{4} x+\cos ^{4} x\right)}$
is equal to:
$x \tan \left(\frac{y}{x}\right) d y=\left(y \tan \left(\frac{y}{x}\right)-x\right) d x,-1 \leq x \leq 1, y\left(\frac{1}{2}\right)=\frac{\pi}{6}$
Then the area of the region bounded by the curves $x=0, x=\frac{1}{\sqrt{2}}$ and $y=y(x)$ in the upper half plane is :