Find second derivative of $ \space g(x)=\int_{0}^{x^{2}} e^{t} dt \space$ and solve for $x = 2$

Question

I am trying to solve the problem

$$ \text { Let } g(x)=\int_{0}^{x^{2}} e^{t} d t $$ $$ \text { Find } g^{\prime \prime}(2) $$

I keep on getting $18e^4$ but someone else told me they got $4e^4$. This is my work, is there is a problem I'm not seeing with it? Thanks in advance! $$ \begin{aligned} g(x) &=\int_{0}^{x^{2}} e^{t} d t \\ g^{\prime}(x) &=e^{x} \cdot 2 x \\ g^{\prime \prime}(x) &=2 e^{x^{2}}+2 x \cdot 2 x e^{x^{2}} \\ &=2 e^{x^{2}}\left(1+2 x^{2}\right) \\ g^{\prime \prime}(2) &=18 e^{4} \end{aligned} $$

Answer 1

I’ll go through the working at a slightly slower pace to the other answers in order to make the step by step work a little clearer.

Step 1

We have that: $g(x) =\int_{0}^{x^{2}} e^{t} d t$

Note that the Chain Rule and Fundamental Theorem of Calculus say that we can directly substitute $x^2$ for $t$ and multiply by the derivative of $x^2$ in order to find the first derivative of $g$.

This gives us $g’(x)=2xe^{x^2}$

Step 2

By the product rule, we are able to differentiate $g’$ in order to give us an expression for $g’’$.

The product rule states that if we are able to decompose a function into two separate functions of the same variable, then we can differentiate the first function and multiply this derivative by the second function. And then add this to the derivative of the second function multiplied by the first function.

If we take $2x$ to be our first function and $e^{x^2}$ to be our second, then the derivatives of these with respect to $x$ are $2$ and $2xe^{x^2}$ respectively.

And so by the product rule, the second derivative is:

$$g’’(x)=4x^2e^{x^2}+2e^{x^2}$$

Step 3

If we substitute in $x=2$, then this gives us

$$g’’(2)=16e^4+2e^4=18e^4$$

And so you are quite right, that the answer will be $18e^4$ for this question.

Answer 2

Well, I got this:

$$ \begin{aligned} g(x) &=\int_{0}^{x^{2}} e^{t} dt = g(x) = e^{x^2} - 1 \\ g'(x) &= 2xe^{x^2} \\ g''(x) &= 2e^{x^2} + 4x^2e^{x^2}\\ g''(x) &= 2e^{2^2} + 4(2)^2e^{(2)^2} = 18e^{4} \end{aligned} $$

You're welcome.

Edit: Sorry, I missed the second derivative. I have edited my answer to include this now.

Answer 3

$$g(x) = \int_0^{x^2} e^t\ dt = e^{x^2} - 1$$

Now, $$g'(x) = 2x e^{x^2}$$

Differentiating it again, $$g''(x) = 2e^{x^2} + 2x\cdot e^{x^2}\cdot 2x \implies g''(x) = 2e^{x^2} + 4x^2e^{x^2}$$

Putting $x=2$ we have: $$g''(2) = 2e^{2^2} + 4(2)^2\cdot e^{2^2} \implies g''(2) = 2e^4 + 16 e^4 =\boxed{18e^4}$$