JEE Advance - Mathematics (1994 - No. 44)
If $$y = {\left( {\sin x} \right)^{\tan x}},$$ then $${{dy} \over {dx}}$$ is equal to
$${\left( {\sin x} \right)^{\tan x}}\left( {1 + {{\sec }^2}x\,\log \,\sin \,x} \right)$$
$$\tan x{\left( {\sin x} \right)^{\tan x - 1}}.\cos x$$
$${\left( {\sin x} \right)^{\tan x}}{\sec ^2}x\,\log \,\sin \,x$$
$$\tan x{\left( {\sin x} \right)^{\tan x - 1}}$$