Mathematics - Differentiation Question with Solution | TestHub

For $f(x)=(\tan x{)}^{\cot x}$, $\ln f(x)=\cot x\ln (\tan x)$. Differentiating, $\frac{f^{\prime }(x)}{f(x)}=-{\csc }^{2}x\ln (\tan x)+\cot x\frac{{\sec }^{2}x}{\tan x}={\csc }^{2}x(1-\ln (\tan x))$. At$x=\pi /4$, $f(\pi /4)=1$. $f^{\prime }(\pi /4)=1\cdot ({\csc }^2(\pi /4))(1-\ln (\tan (\pi /4)))=2(1-0)=2$. For $g(x)=(\cot x{)}^{\tan x}$, $\ln g(x)=\tan x\ln (\cot x)$. Differentiating, $\frac{g^{\prime }(x)}{g(x)}={\sec }^{2}x\ln (\cot x)+\tan x\frac{-{\csc }^{2}x}{\cot x}={\sec }^{2}x(\ln (\cot x)-1)$. At$x=\pi /4$, $g(\pi /4)=1$. $g^{\prime }(\pi /4)=1\cdot ({\sec }^2(\pi /4))(\ln (\cot (\pi /4))-1)=2(0-1)=-2$. P. $f^{\prime }(\pi /4)=2$. Q.$g^{\prime }(\pi /4)=-2$. R.${\frac{d}{dx}(f(x)+g(x))|}_{x=\pi /4}=f^{\prime }(\pi /4)+g^{\prime }(\pi /4)=2+(-2)=0$. S.${\frac{d}{dx}(f(x)g(x))|}_{x=\pi /4}=f^{\prime }(\pi /4)g(\pi /4)+f(\pi /4)g^{\prime }(\pi /4)=(2)(1)+(1)(-2)=0$. Thus, P-(2), Q-(3), R-(1), S-(1).