Derivative and Integral problems - PLEASE HELP! :)

**RedSpades** · October 14th 2008, 04:35 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** · October 14th 2008, 04:40 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** · October 14th 2008, 04:42 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** · October 14th 2008, 04:44 PM

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

**skeeter** · October 14th 2008, 04:44 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** · October 14th 2008, 04:45 PM

wow my teacher does it differently

**bigton** · October 14th 2008, 04:48 PM

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

**RedSpades** · October 14th 2008, 04:53 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** · October 14th 2008, 05:04 PM

Can you guys go a little further please

**RedSpades** · October 14th 2008, 05:17 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** · October 14th 2008, 05:21 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** · October 14th 2008, 05:23 PM

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

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

is this right

**RedSpades** · October 14th 2008, 05:24 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.

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