Question

How do you find the derivatives?
int_{0}^{4t^2}sqrt(1-3x^3)dt
int_{3}^{lnx}tan(t) dt

Answer 1

The fundamental theorem of calculus \[
F(x) = \int_a^x f(t)dt \implies F'(x) = f(x)
\]

Answer 2

for the first one, do you mean to find F'(x) when \(\large F(x)=\int_{0}^{4x^2}\sqrt{1-3t^3}dt \) ???

Answer 3

Yes. That's what I mean.
I don't really know how to use that theorem.

Answer 4

\[ \large
F(x)=\int_{0}^{4x^2}\sqrt{1-3t^3}dt \
\]So we can't quite use the fundamental theorem yet, because of that nasty \(4x^2\).
Let's suppose that: \(y=4x^2\) and \(F(x) = G(y) \).
Using the fundamental theorem, we get: \[ \large G(y)=\int_{0}^{y}\sqrt{1-3t^3}dt \implies G'(y) = \sqrt{1-3y^3}
\]Now, consider that \(F(x) = G(y) = G(y(x))\). If we use the chain rule, we get: \[ \large
F'(x) = G'(y(x)) \cdot y'(x)
\]

Answer 5

We know \[ \large
G'(y(x)) = \sqrt{1-3[y(x)]^3} =\sqrt{1-3[4x^2]^3} = \sqrt{1-192x^6} \\ \large
y(x) = 4x^2 \\ \large y'(x) = 8x \\ \large
F'(x) = \sqrt{1-192x^6} \cdot 8x
\]

Answer 6

For problem 2, we get: \[
G'(y) = \tan(y) \\
y'(x) = \frac{1}{x} \\
F'(x) = \frac{\tan(\ln(x))}{x}
\]