Derivative and Integral problems - PLEASE HELP! :)

**RedSpades** · Oct 14th 2008, 03:35 PM

Hello guys.

I need quick help on these two problems from my math packet. One of them having to do with Derivatives and the other having to do with Integrals.

Here are the problems I need the most help in:

$f(x)= \sqrt{ e^{7x} + e^{-7x}}$

and

$\int\frac{e^{\frac{4}{x^{2}}}}{6x^{3}}dx$

I really appreciate the help!

Thank you all.

**mr fantastic** · Oct 14th 2008, 03:40 PM

Originally Posted by RedSpades

Hello guys.

I need quick help on these two problems from my math packet. One of them having to do with Derivatives and the other having to do with Integrals.

Here are the problems I need the most help in:

$f(x)= \sqrt{ e^{7x} + e^{-7x}}$

Mr F says: Use the chain rule. Let ${\color{red}u = e^{7x} + e^{-7x}}$ .

and

$\int\frac{e^{\frac{4}{x^{2}}}}{6x^{3}}dx$

Mr F says: Make the sustitution ${\color{red}u = \frac{4}{x^2}}$ .

I really appreciate the help!

Thank you all.

..

**bigton** · Oct 14th 2008, 03:42 PM

I am trying to follow but i am still a little lost. This site is great tho this is what im learning in class. Well trying

**bigton** · Oct 14th 2008, 03:44 PM

because u r finding f ' (x) not the integral

**skeeter** · Oct 14th 2008, 03:44 PM

$f(x) = (e^{7x} + e^{-7x})^{\frac{1}{2}}$

straight-forward ... take the derivative using the chain rule.

$\int \frac{e^{\frac{4}{x^2}}}{6x^3} \, dx$

substitution ...

let $u = \frac{4}{x^2}$

$du = -\frac{8}{x^3} \, dx$

$-\frac{1}{6 \cdot 8} \int e^{\frac{4}{x^2}} \cdot \left(-\frac{8}{x^3}\right) \, dx$

**bigton** · Oct 14th 2008, 03:45 PM

wow my teacher does it differently

**bigton** · Oct 14th 2008, 03:48 PM

but for the first question how do u slove the e's in side

**RedSpades** · Oct 14th 2008, 03:53 PM

Thanks for the help, guys.

I'm going to try to figure out the one with the chain rule. We just started using that rule so I might be back for some help.

**theflyingcow** · Oct 14th 2008, 04:04 PM

Can you guys go a little further please

**RedSpades** · Oct 14th 2008, 04:17 PM

Originally Posted by RedSpades

Hello guys.

$f(x)= \sqrt{ e^{7x} + e^{-7x}}$

So guys, am I doing this right?

$<br /> f(x)= (e^{7x} + e^{-7x})^{\frac{1}{2}}$

$f'(x)= \frac{1}{2}(e^{7x} + e^{-7x})^{-\frac{1}{2}}$

Now I was thinking that the e's in the parathesis can be differentiated but I always get confused with the e's and the fact that the exponents have the variable 'x'. How do I continue from here?

**mr fantastic** · Oct 14th 2008, 04:21 PM

Originally Posted by RedSpades

So guys, am I doing this right?

$<br /> f(x)= (e^{7x} + e^{-7x})^{\frac{1}{2}}$

$f'(x)= \frac{1}{2}(e^{7x} + e^{-7x})^{-\frac{1}{2}} {\color{red}\left(7 e^{7x} - 7 e^{-7x}\right)}$ Mr F edits: You need the stuff in red.

Now I was thinking that the e's in the parathesis can be differentiated but I always get confused with the e's and the fact that the exponents have the variable 'x'. How do I continue from here?

Chain rule: $\frac{dy}{dx} = \frac{dy}{du} \, \frac{du}{dx}$ .

So the stuff in red is needed. It's the 'du/dx bit' of the chain rule. Your answer only has the 'dy/du' bit ....

**uniquereason81** · Oct 14th 2008, 04:23 PM

(e^(7x) * 7 + e^(-7x) * -7)^1/2

7e^(7x )+ -7e^(-7x)

is this right

**RedSpades** · Oct 14th 2008, 04:24 PM

Originally Posted by mr fantastic

Chain rule: $\frac{dy}{dx} = \frac{dy}{du} \, \frac{du}{dx}$ .

So the stuff in red is needed. It's the 'du/dx bit' of the chain rule. Your answer only has the 'dy/du' bit ....

Ohhhh! I forgot to do the rest of that. I knew I was missing a part of the rule.
Thanks Mr. Fantastic.

Thread: Derivative and Integral problems - PLEASE HELP! :)

LinkBack

Thread Tools

Display

Derivative and Integral problems - PLEASE HELP! :)

Similar Math Help Forum Discussions

Directional Derivative & Triple integral problems

Two derivative problems

2 derivative problems

derivative problems

Various Derivative Problems

Search Tags