TITLE: How can I calculate the derivative of $g(z)=\int_{z^2}^{z^3}\frac{1}{\sqrt{1+t^2}} dt$?
QUESTION [11 upvotes]: How can I calculate the derivative of $$g(z)=\int_{z^2}^{z^3}\frac{1}{\sqrt{1+t^2}} dt$$
Is there anything I can do apart from calculating the integral and then derivate?

REPLY [29 votes]: $$f:t\mapsto \frac {1}{\sqrt {1+t^2}} $$
is continuous at R, thus
$$F:x\mapsto \int_0^xf (t)dt $$ is differentiable at R and
$$F'(x)=f (x) $$
your function is
$$g (z)=F (z^3)-F (z^2) $$
it is differentiable at R and
$$g'(z)=3z^2f (z^3)-2zf (z^2) $$
$$=\frac {3z^2}{\sqrt {1+z^6}}-\frac {2z}{\sqrt {1+z^4}} $$