Mathematics - Differentiation Question with Solution | TestHub

Since $f(x)=\sin \left(\frac{2\times 3^x}{1+9^x}\right)$ Suppose $3^x=\tan t$ $$\begin{array}{rl}& \Rightarrow f(x)={\sin }^{-1}\left(\frac{2\tan t}{1+{\tan }^{2}t}\right)\\ & ={\sin }^{-1}(\sin 2t)=2t=2{\tan }^{-1}(3x)\\ & \text{ So, }{f}^{\prime }(x)=\frac{2}{1+{\left(3^x\right)}^2}\times 3^x\cdot {\log }_{e}3\\ & \therefore f^{\prime }\left(-\frac{1}{2}\right)=\frac{2}{1+{\left(3^{-\frac{1}{2}}\right)}^2}\times 3^{-\frac{1}{2}}\cdot {\log }_{e}3\\ & =\frac{1}{2}\times \sqrt{3}\times {\log }_{e}3=\sqrt{3}\times {\log }_e\sqrt{3}\end{array}$$