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OpenStudy (anonymous):
Let \(u = -3/t\).
OpenStudy (anonymous):
\[
y = 3e^{u}
\]
OpenStudy (anonymous):
\[
\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}
\]
OpenStudy (anonymous):
Can you find these two derivatives?
OpenStudy (anonymous):
there is no u substitutions when finding the derivative
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OpenStudy (anonymous):
What is \(dy/du\)?
OpenStudy (anonymous):
integrals
OpenStudy (anonymous):
I am not doing an integral.
OpenStudy (anonymous):
\[
y = 3e^u
\]What is the derivative?
OpenStudy (anonymous):
Im not sure with that u
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OpenStudy (anonymous):
Well, I want \(dy/du\) so you are differentiating with respect to \(u\). It is an easy one.
OpenStudy (anonymous):
Ok, can you walk me through this?
OpenStudy (anonymous):
Okay. Well \[
y=3e^{u}\implies \frac{dy}{du}=3e^u
\]Does this make sense?
OpenStudy (anonymous):
so u=-3t^-1?
OpenStudy (anonymous):
Yes, just go with me for a moment. \[
u = -\frac{3}{t}
\]
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OpenStudy (anonymous):
We need to find \(du/dt\) now.
OpenStudy (anonymous):
du=3t^-2
OpenStudy (anonymous):
Ok, so now we say: \[
\frac{dy}{dt} = \frac{dy}{du}\frac{du}{dt} = 3e^u\frac{3}{t^2}
\]Now, we can change \(u\) back into terms of \(t\): \[
\frac{dy}{dt}=3e^{-3/t}\frac{3}{t^2}
\]
OpenStudy (anonymous):
Does this make sense?
OpenStudy (kainui):
@Bosse15 Trust wio, he's just doing the Chain Rule, it just looks a little different than what you're probably used to.
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OpenStudy (anonymous):
yes it does
OpenStudy (anonymous):
Help?
OpenStudy (anonymous):
\[\int\limits_{?}^{?}xe ^{^{^{-3x ^{2}}}}dx\]
OpenStudy (anonymous):
Help with what?
OpenStudy (anonymous):
This time let \(u = -3x^2\)
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OpenStudy (anonymous):
\[
du = d(-3x^2) =(-3x^2)'dx = -6xdx
\]
OpenStudy (anonymous):
ok, then du=-6xdx
OpenStudy (anonymous):
-1/6 e^-3x^2+C?
OpenStudy (anonymous):
\[
\int e^{u}xdx
\]We can do this: \[
=\frac{-6}{-6}\int e^{u}xdx = -\frac{1}{6}\int e^{u}(-6x)dx
\]
OpenStudy (anonymous):
Then it becomes: \[
-\frac{1}{6}\int e^u\;du
\]
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OpenStudy (anonymous):
Oh, your answer is correct.
OpenStudy (anonymous):
Cool! lol thanks, could you stay on here just to make sure i can finish the rest of these up?
OpenStudy (anonymous):
\[\int\limits_{}^{}\frac{ e ^{\frac{ 1 }{ x }} }{ x ^{2} }\]
OpenStudy (anonymous):
dx\
OpenStudy (anonymous):
try using wolfram alpha
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OpenStudy (anonymous):
never heard of that
OpenStudy (kainui):
www.wolframalpha.com It's really helpful for checking the answers to calculus problems.
OpenStudy (anonymous):
\[\int\limits_{}^{}\frac{ e ^{2x} }{ e ^{2x}+1 }dx\]
OpenStudy (anonymous):
@wio
OpenStudy (kainui):
Wait I'm overcomplicating this. This is really quite a simple u-subsitution problem.
e^2x+1=u