Question

If f(x)=tan^2(2x+1) then find f'(-1).

Answer 1

Chain rule!\[\frac {d}{dx} \left[ f(x)=\tan^2(2x+1) \right]\]\[\frac {df}{dx} = 2\tan(2x+1)* \frac {d}{dx} \tan (2x + 1)\]\[\frac {df}{dx} = 2\tan(2x+1)* \sec^2 (2x + 1) * \frac {d}{dx} (2x + 1)\]\[\frac {df}{dx} = 4 \tan (2x+1) \sec^2(2x+1)\]\[f'(-1) = 4 \tan (2*-1+1) \sec^2(2*-1+1)\]\[f'(-1) = 4 \tan (-1) \sec^2 (-1) = -4 \tan (1) \sec^2 (1) \approx -21.3397\]

Answer 2

First differentiate the power, then the function, then the inner function.

Answer 3

thanks

Answer 4

No problem :)