Differentiate f(x)=tan(x^2+1).

We'll write tan(x^2 + 1) = sin(x^2 + 1)/cos (x^2 +
1)

We'll apply quotient rule and chain
rule:

f'(x) = [sin(x^2 + 1)]'*cos (x^2 + 1) - sin(x^2 +
1)*[cos (x^2 + 1)]'/[cos (x^2 + 1)]^2

f'(x) = 2x*cos (x^2 +
1)*cos (x^2 + 1) + 2xsin(x^2 + 1)*sin(x^2 + 1)/[cos (x^2 +
1)]^2

f'(x) = 2x{[cos (x^2 + 1)]^2 + [sin (x^2 +
1)]^2}/[cos (x^2 + 1)]^2

But, from Pythagorean identity,
[cos (x^2 + 1)]^2 + [sin (x^2 + 1)]^2 =
1

f'(x) = 2x/[cos (x^2 +
1)]^2