Answer :
Answer:
Step-by-step explanation:
[tex]f(x)=tan^{-1} 2x\\let f(x)=y\\y=tan^{-1}2x\\tan y=2x\\diff. w.r.t.x\\sec^2y(\frac{dy}{dx} )=2\\\frac{dy}{dx} =\frac{2}{sec^2 y} =\frac{2}{1+tan^2y} =\frac{2}{1+(2x)^2} \\f'(x)=\frac{2}{1+4x^2}[/tex]
Answer:
Step-by-step explanation:
[tex]f(x)=tan^{-1} 2x\\let f(x)=y\\y=tan^{-1}2x\\tan y=2x\\diff. w.r.t.x\\sec^2y(\frac{dy}{dx} )=2\\\frac{dy}{dx} =\frac{2}{sec^2 y} =\frac{2}{1+tan^2y} =\frac{2}{1+(2x)^2} \\f'(x)=\frac{2}{1+4x^2}[/tex]