Question

what is the differentiate of 4tanx/1-tan^x

Answer 1

Let \[f(x) = 4 \tan{x}, g(x) = 1 - \tan^2{x}\] Then \[\frac{d}{dx} \frac{4 \tan{x}}{1 - \tan^2{x}} = \frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f'(x) g(x) - g'(x) f(x)}{g(x)^2} =\]\[ \frac{4 \sec^2{x} \cdot (1 - \tan^2{x}) + 8 \tan^2{x} \sec^2{x} }{(1 - \tan^2{x})^2} = \cdots = 4 \sec^2{2x}\]