I tried to solve this question. It is a multiple choice question.
First I took the derivative of tan(ax+b) which is $\sec^2(ax+b)$ but If I take the derivative with respect to x it will be a $\sec^2(ax+b)$.
what I find confusing is $d/d \tan(ax+b) = \sec^2 (ax+b) d/d \tan(ax+b) *(ax+b)$
I can't really understand this problem, I have been trying many times.
Best Answer
The derivative is $$\frac{d}{dx}[\tan(ax+b)] = \sec^2(ax+b)\cdot\frac{d}{dx}[ax+b] = a\sec^2(ax+b),$$ where the second equality invokes the chain rule.
Edit: Recall that $$\frac{dg}{df} = \frac{dg}{dx}\cdot\frac{dx}{df} = \frac{g'(x)}{f'(x)}$$ So, in our case $g(x) =\tan(ax+b)$. But $f(x) = \tan(ax+b)$ also. Therefore $$\frac{g'(x)}{f'(x)} = \frac{a\sec^2(ax+b)}{a\sec^2(ax+b)}=1.$$