Solving a question using the fundamental theorem of calculus I'm using the fundamental theorem of calculus to try to solve the following problem...
If $f(x)=\int_0^{x^2}t^4 \, dt$, then what is $f'(x)$?
Using the fundamental theorem of Calculus I get $f'(x)= x^8$
Am i doing this right or no?
 A: By the fundamental theorem of calculus $\displaystyle \int_{0}^{x^2}t^4dt = G(x^2)-G(0)$ for a differentiable function $G$. This implies that $f'(x)=\frac{d}{dx}(G(x^2)-G(0))=2xG'(x^2)$. Since, $G'(x)=x^4$ we obtain $f'(x)=2x^9$. By the way, you can check your work by just evaluating the integral directly.
A: Let $g(t)=t^4$ and $G$ -antiderivative of $g$ $G'(t)=g(t)$. Then by Fundamental Theorem of Calculus:
$$f(x)=\int_{0}^{x^2} g(t) dt=G(x^2)-G(0)$$
Derive both sides (right side using chain rule):
$$f'(x)=(G(x^2)-G(0))=(x^2)'G'(x^2)=2xg(x^2)=2x \cdot (x^{2})^4=2x^{9}$$
A: recall that after integrating that $t$ disappears.
first, find $f$
$$f(x) = \int_0^{x^2}t^4dt = \frac{t^5}{5}\bigg|_0^{x^2}=\frac{x^{10}}{5}-0=\frac{x^{10}}{5}$$
then $f'$
$$f'(x)=\frac{d}{dx} \bigg(\frac{x^{10}}{5}\bigg)=\frac{10x^9}{5}=2x^9$$
A: Just put $u=x^{2}$ and note that $$y=f(x) = \int_{0}^{u}t^{4}\,dt$$ and then by FTC we have $$\frac{dy}{du} = u^{4}$$ and then by chain rule we have $$f'(x) = \frac{dy}{dx} = \frac{dy}{du} \cdot\frac{du}{dx} = u^{4}\cdot 2x = 2x^{9}$$
