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