Use the fundamental theorem of calculus to find the derivative of given functions

Find the derivative $f'(x)$ for each of the following (without attempting to evaluate the integrals)

For all parts of the problem let

$A(x) = \int_0^x (1+t^2)^{-3} \, dt.$

From the fundamental theorem of calculus we then know

$A'(x) = (1+x^2)^{-3}.$

Now, $f(x) = A(x)$ so
$f'(x) = A'(x) = (1+x^2)^{-3}.$
In this case $f(x) = A(x^2)$ so
$f'(x) = (A(x^2))' = (2x)A'(x^2) = (2x)(1+x^4)^{-3}.$
Now, we rewrite the integral,
$f(x) = \int_{x^3}^{x^2} (1+t^2)^{-3} \, dt = \int_0^{x^2} (1+t^2)^{-3} - \int_0^{x^3} (1+t^2)^{-3} \, dt.$

Therefore,

$f(x) = A(x^2) - A(x^3) \quad \implies \quad f'(x) = (A(x^2))' - (A(x^3))'.$

From above we know

$\begin{align*} (A(x^2))' = (2x) A'(x^2) &= (2x)(1+x^4)^{-3} \\ (A(x^3))' = (3x^2) A'(x^3) &= (3x^2)(1+x^6)^{-3}. \end{align*}$

Therefore,

$f'(x) = (2x)(1+x^4)^{-3} - 3x^2(1+x^6)^{-3}.$