Answer
$$\frac{{dy}}{{dx}} = \frac{{3{{\sec }^2}}}{{16}}x - 8{\csc ^2}2x + 5\csc x\cot x - 2{e^{ - 2x}}$$
Work Step by Step
$$\eqalign{
& y = 3\tan \left( {\frac{1}{4}x} \right) + 4\cot 2x - 5\csc x + {e^{ - 2x}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}\left[ {3\tan \left( {\frac{1}{4}x} \right) + 4\cot 2x - 5\csc x + {e^{ - 2x}}} \right] \cr
& {\text{sum rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = {D_x}\left[ {3\tan \left( {\frac{1}{4}x} \right)} \right] + {D_x}\left[ {4\cot 2x} \right] - {D_x}\left[ {5\csc x} \right] + {D_x}\left[ {{e^{ - 2x}}} \right] \cr
& \frac{{dy}}{{dx}} = 3{D_x}\left[ {\tan \left( {\frac{1}{4}x} \right)} \right] + 4{D_x}\left[ {\cot 2x} \right] - 5{D_x}\left[ {\csc x} \right] + {D_x}\left[ {{e^{ - 2x}}} \right] \cr
& {\text{use the chain rule and the differentiation rule for trigonometric}} \cr
& {\text{functions}}{\text{, then}} \cr
& \frac{{dy}}{{dx}} = 3{\sec ^2}\left( {\frac{1}{4}x} \right){D_x}\left[ {\frac{1}{4}x} \right] + 4\left( { - {{\csc }^2}2x} \right){D_x}\left[ {2x} \right] - 5\left( { - \csc x\cot x} \right) - 2{e^{ - 2x}} \cr
& \frac{{dy}}{{dx}} = \frac{{3{{\sec }^2}}}{4}x\left( {\frac{1}{4}} \right) + 4\left( { - {{\csc }^2}2x} \right)\left( 2 \right) + 5\csc x\cot x - 2{e^{ - 2x}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{3{{\sec }^2}}}{{16}}x - 8{\csc ^2}2x + 5\csc x\cot x - 2{e^{ - 2x}} \cr} $$
