Second Fundamental theorem of Calculus find f'(x)

[goob94]

Using Second Fundamental theorem of Calculus find f'(x) for f(x)=∫√(t^2+1) dt for [2x, 3x^2]

 

[MarkFL]

The anti-derivative form of the FTOC states:

\(\displaystyle \displaystyle \int_a^b f(x)\,dx=F(b) - F(a)\) where \(\displaystyle \dfrac{dF}{dx}=f(x)\) and so:

\(\displaystyle \displaystyle \frac{d}{dx}\left(\int_{2x}^{3x^2}\sqrt{t^2+1}\,dt \right)=\)

\(\displaystyle \displaystyle \frac{d}{dx}\left(F(3x^2)-F(2x) \right)=\)

\(\displaystyle \displaystyle \frac{d}{dx}(F(3x^2))-\frac{d}{dx}(F(2x))=\)

Can you now complete this using the chain rule?

 

[goob94]

The anti-derivative form of the FTOC states:

\(\displaystyle \displaystyle \int_a^b f(x)\,dx=F(b) - F(a)\) where \(\displaystyle \dfrac{dF}{dx}=f(x)\) and so:

\(\displaystyle \displaystyle \frac{d}{dx}\left(\int_{2x}^{3x^2}\sqrt{t^2+1}\,dt \right)=\)

\(\displaystyle \displaystyle \frac{d}{dx}\left(F(3x^2)-F(2x) \right)=\)

\(\displaystyle \displaystyle \frac{d}{dx}(F(3x^2))-\frac{d}{dx}(F(2x))=\)

Can you now complete this using the chain rule?

I don't really understand how to use the chain rule, can you continue it?

 

[MarkFL]

\(\displaystyle \displaystyle \frac{d}{dx}\left(F(u(x) \right)=\frac{dF}{du}\cdot\frac{du}{dx}=f(u)\cdot \frac{du}{dx}\)

Now, in our case, we have \(\displaystyle f(t)=\sqrt{t^2+1}\) and for:

the first term: \(\displaystyle u(x)=3x^2\)

the second term: \(\displaystyle u(x)=2x\)

So, what do you find?

 

[pka]

Using Second Fundamental theorem of Calculus find f'(x) for f(x)=∫√(t^2+1) dt for [2x, 3x^2]

I like to state the Second Fundamental theorem of Calculus
a different way.

Suppose that each of \(\displaystyle f~\&~g\) is a differentiable function and \(\displaystyle \Phi (x) = \int_{g(x)}^{f(x)} {H(t)dt} \) then \(\displaystyle \Phi '(x) = f'(x)H(f(x)) - g'(x)H(g(x))\).

In your problem is \(\displaystyle \Phi (x) = \int_{2x}^{3x^2 } {\sqrt {t^2 + 1} dt} \).

So \(\displaystyle f(x)=3x^2,~g(x)=2x,~\&~H(t)=\sqrt{t^2+1}\)

 

[goob94]

Using Second Fundamental theorem of Calculus find f'(x) for f(x)=∫√(t^2+1) dt for [2x, 3x^2]

I like to state the Second Fundamental theorem of Calculus
a different way.

Suppose that each of \(\displaystyle f~\&~g\) is a differentiable function and \(\displaystyle \Phi (x) = \int_{g(x)}^{f(x)} {H(t)dt} \) then \(\displaystyle \Phi '(x) = f'(x)H(f(x)) - g'(x)H(g(x))\).

In your problem is \(\displaystyle \Phi (x) = \int_{2x}^{3x^2 } {\sqrt {t^2 + 1} dt} \).

So \(\displaystyle f(x)=3x^2,~g(x)=2x,~\&~H(t)=\sqrt{t^2+1}\)

how do i find h(f(x)) and h(g(x))

it would be then (6x)h(f(x))-2(h(g(x))

 

[pka]

how do i find h(f(x)) and h(g(x))

it would be then (6x)h(f(x))-2(h(g(x))

Are you actually telling us (and the world) that you cannot find \(\displaystyle H(f(x))\) but you think you can do this question?

You don't know that if \(\displaystyle H(t)=\sqrt{t^2+1}~\&~f(x)=3x^2\) then \(\displaystyle H(f(x))=\sqrt{9x^4+1}\).

If not, why are you doing this question?

 

[MarkFL]

\(\displaystyle \displaystyle \frac{d}{dx}\left(F(u(x) \right)=\frac{dF}{du}\cdot\frac{du}{dx}=f(u)\cdot \frac{du}{dx}\)

Now, in our case, we have \(\displaystyle f(t)=\sqrt{t^2+1}\) and for:

the first term: \(\displaystyle u(x)=3x^2\)

the second term: \(\displaystyle u(x)=2x\)

So, what do you find?

This would give us:

\(\displaystyle \displaystyle f(3x^2)\cdot \frac{d}{dx}(3x^2)-f(2x)\cdot \frac{d}{dx}(2x)=\)

\(\displaystyle \displaystyle \sqrt{(3x^2)^2+1}\cdot6x-\sqrt{(2x)^2+1}\cdot2=\)

\(\displaystyle \displaystyle 6x\sqrt{9x^4+1}-2\sqrt{4x^2+1}\)

Does this make sense...please don't hesitate to ask if anything is unclear.

 

[goob94]

This would give us:

\(\displaystyle \displaystyle f(3x^2)\cdot \frac{d}{dx}(3x^2)-f(2x)\cdot \frac{d}{dx}(2x)=\)

\(\displaystyle \displaystyle \sqrt{(3x^2)^2+1}\cdot6x-\sqrt{(2x)^2+1}\cdot2=\)

\(\displaystyle \displaystyle 6x\sqrt{9x^4+1}-2\sqrt{4x^2+1}\)

Does this make sense...please don't hesitate to ask if anything is unclear.

yes, thank you