Question

Differentiate the function w.r.t.x. : $e^{2x}. \tan^{-1}2x$

Answer 1

y=e^2x . tan^-1 (2x)
diff. w.r.t.x
: . dy/dx
= [e^2x . 2/(4x^2+1)] +[ tan^-1 (2x) . 2e^2x]
= 2e^2x[1/{(4x)^2+1} + tan^-1 (2x)]

Answer 2

Answer:

$\frac{d e^{2x} .tan^{-1}2x }{dx}=2 e^{2x}( tan^{-1}2x+\frac{1}{1+4x^{2} } )$

Step-by-step explanation:

For differentiating the given function we must use U.V form of differentiation

$\frac{d U.V}{dx}=U.\frac{dV}{dx}+V. \frac{dU}{dx}$

here

$U = e^{2x} \\ \\ V=tan^{-1} 2x$

$\frac{d e^{2x} .tan^{-1}2x }{dx}=e^{2x}.\frac{dtan^{-1}2x}{dx}+tan^{-1}2x. \frac{de^{2x}}{dx}$

As we know that

$\frac{d e^{f(x)} }{dx} =e^{f(x)}.\frac{{f(x)}}{dx}\\ \\ \frac{d tan^{-1}f(x) }{dx} =\frac{1}{1+x^{2} } .\frac{{f(x)}}{dx}$

$\frac{d e^{2x} .tan^{-1}2x }{dx}=e^{2x}.\frac{1}{1+4x^{2} } (\frac{d2x}{dx})+tan^{-1}2x. e^{2x}.(\frac{d2x}{dx})\\ \\ \\ =\frac{2 e^{2x}}{1+4x^{2} } +2e^{2x}.tan^{-1}2x\\ \\ \\ =2 e^{2x}( tan^{-1}2x+\frac{1}{1+4x^{2} } )$