Question

Find the derivative of the following function:

$f(x) = \int_{1}^{x^2} 2t \,dt$

Answer

To find the derivative of the function, we will apply the Fundamental Theorem of Calculus. According to the theorem, if $f(t) = \int_{a}^{g(t)} h(u) \,du$, then $f'(t) = h(g(t)) \cdot g'(t)$.

In the given function, $f(x) = \int_{1}^{x^2} 2t \,dt$. Here, $h(u) = 2u$ and $g(t) = t^2$. So,

$f'(x) = h(g(x)) \cdot g'(x)$

To find $h(g(x))$, we substitute $g(x) = x^2$ into $h(u) = 2u$:

$h(g(x)) = 2 \cdot (x^2)$

Next, we find the derivative of $g(x)$ using the chain rule. The derivative of $g(x) = x^2$ with respect to $x$ is:

$g'(x) = 2x$

Finally, we substitute these values back into the equation to find the derivative of $f(x)$:

$f'(x) = 2 \cdot (x^2) \cdot (2x)$

Simplifying the expression:

$f'(x) = 4x^3$

Therefore, the derivative of the function $f(x) = \int_{1}^{x^2} 2t \,dt$ is $f'(x) = 4x^3$.