# Thread: Fund. Thm of Calc Question

1. ## Fund. Thm of Calc Question

Let $\displaystyle G(x)=\int^{2x}_{3x}\;f(t)\:dt$, where $\displaystyle f(t)=\int^{2-t}_{2t-5}\sqrt{1+u^4}\;du$. Then $\displaystyle G''(1)=$

2. Originally Posted by polymerase
Let $\displaystyle G(x)=\int^{2x}_{3x}\;f(t)\:dt$, where $\displaystyle f(t)=\int^{2-t}_{2t-5}\sqrt{1+u^4}\;du$. Then $\displaystyle G''(1)=$
As a start, $\displaystyle G^{'}(x) = 2 f(2x) - 3 f(3x)$.

So now you need to differentiate $\displaystyle 2 f(2x) - 3 f(3x)$ with respect to x, where you know the rule for f ........

3. Originally Posted by mr fantastic
As a start, $\displaystyle G^{'}(x) = 2 f(2x) - 3 f(3x)$.

So now you need to differentiate $\displaystyle 2 f(2x) - 3 f(3x)$ with respect to x, where you know the rule for f ........
what do you do now?

4. Originally Posted by polymerase
what do you do now?
Indeed, what do you do now?

At the risk of sounding arrogant, I would have thought it was reasonably obvious that now you get expressions for f(2x) and f(3x) using the given rule $\displaystyle f(t)=\int^{2-t}_{2t-5}\sqrt{1+u^4}\;du$.

Then you construct the expression 2f(2x) - 3f(3x). Then you differentiate it.