Find this Calculus question Confusing

Question

Given that $F=\int _{2x}^{x²} f(x)$\, dx 

What is the value of $F'(x)$

I thought this should just Be $f(x²)-f(2x)...$But my Class' toppper got; $2xf(x²)+2f(2x)...$Think he did something about implicit differentiation.

Can anyone please help me out here?

Answer 1

Suppose you know an anti-derivative $G$ for $f$. Then $G' = f$. So then

$$F(x) = \int_{2x}^{x^2} f(t) \; dt = G(x^2) - G(2x).$$

Then by chain rule

$$F'(x) = G'(x^2)(2x) - G'(2x)(2) = f(x^2)2x - f(2x)2.$$

Answer 2

I guess you mean this : $$F=\int _{2x}^{x^2} f(t)\, dt$$ Then apply Leibniz rule $$F'(x)=f(x^2)(x^2)'-f(2x)(2x)'$$ $$F'(x)=2xf(x^2)-2f(2x)$$

Answer 3

If $$F=\int_a^x f(x)\ dx$$ then $F'=f(x).$ This means that by the chain rule, if $$F=\int_a^{g(x)} f(x)\ dx$$ we get $F'=f(g(x))g'(x).$ For example this will mean if $$F=\int_a^{x^2} f(x)\ dx$$ we get $F'=f(x^2)2x.$ This explains half the answer. To get the full answer above, split the integral into two parts using the fact that $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx $ and use the chain rule twice.

Answer 4

The notation is rather bad (invalid, even). It should have different variables for the limits and the inside, e.g.

$$F(x)=\int _{2x}^{x^2} f(t)\, dt $$

For simplicity, I'll consider

$$F(x)=\int _{a}^{x^2} f(t)\, dt $$

Then F(x) is the area from $a$ to $x^2$ under the curve of f(t). If you increase $x$ by $h$, then you're now going to $(x+h)^2$, or $x^2+2xh+h^2$. We can ignore the $h^2$ term as $h$ goes to zero. So this is an increase of $2xh$ in your width, at a point where the height is $f((x+h)^2)$. Again, we're taking the limit as $h$ goes to zero, so we can take the height to be $f(x^2)$. Area is height times width, so the additional area is $(2xh)f(x^2)$. The derivative is change in $y$ over change in $x$. $y$, or the area, changed by $(2xh)f(x^2)$, and $x$ changed by $h$. So the derivative is $2xf(x^2)$.

A similar analysis applies to the lower limit of $2x$.