(a)OR(b)

Differentiate $\displaystyle 2^{\cos ^{2} x}$ w. r. t $\displaystyle \cos ^{2} x$.

Let $\displaystyle u=2^{\cos ^{2} x} \Rightarrow \frac{d u}{d x}=2^{\cos ^{2} x}(-2 \cos x \sin x) \log 2$ Let $\displaystyle v=\cos ^{2} x \Rightarrow \frac{d v}{d x}=-2 \cos x \sin x$ Now $\displaystyle \frac{d u}{d v}=\frac{\left(\dfrac{d u}{d x}\right)}{\left(\dfrac{d v}{d x}\right)}=2^{\cos ^{2} x} \log 2$ \[\tan ^{-1}\left(x^{2}+\mathrm{y}^{2}\right)=\mathrm{a}^{2} \Rightarrow x^{2}+\mathrm{y}^{2}=\tan \mathrm{a}^{2} \] Differentiatebothsides wrt $\displaystyle \boldsymbol{x}$, \[\begin{aligned} & 2 x+2 \mathrm{y} \frac{d \mathrm{y}}{d x}=0 \\ & \Rightarrow \frac{d \mathrm{y}}{d x}=-\frac{x}{\mathrm{y}} \end{aligned} \]