Question

find dy/dx given y=definite integral cos(e^t)dt. upperx^2, lower2. show steps/ explain

Answer 1

@SithsAndGiggles

Answer 2

\[y=\int_2^{x^2}\cos\left(e^t\right)~dt\]
In general, the fundamental theorem of calculus states that if \(f(x)=\displaystyle\int_{c}^{u(x)}f(t)~dt\), then its derivative is \(f'(x)=f(u(x))\cdot u'(x)\). (\(c\) is an arbitrary constant.)

Here, \(f(t)=\cos\left(e^t\right)\) and \(u(x)=x^2\).

Answer 3

thnx a million SithsAndGiggles :)

Answer 4

You're welcome!

Answer 5

@SithsAndGiggles sorry for my dummy, I don't see the link between f(u(x) and f(t)

Answer 6

ah no problem, thnx 4 at least trying

Answer 7

@unpi I am not the helper, I want to study from your problem, too.

Answer 8

ok...

Answer 9

that's why I tag the person who I know he can help, not just help you, help me,too

Answer 10

ok, if you got it, show me, please

Answer 11

k, give me a minute

Answer 12

@Loser66, there's no elementary function that is equal to
\[\int\cos\left(e^t\right)~dt\]
But suppose there is, and let \(g(t)=\int\cos\left(e^t\right)~dt\). You can easily replace the \(t\) with \(x\) and it's the same function (the use of \(t\) in my other posts was as a dummy variable).

By the FTC, the definite integral of the function we're working with looks like this, for some limits \(a\) and \(b\):
\[\int_a^b\cos\left(e^t\right)~dt=g(b)-g(a)\]
Let \(b=u(x)\). Then,
\[\int_a^{u(x)}\cos\left(e^t\right)~dt=g(u(x))-g(a)\]
Now take the derivative of both sides with respect to \(x\):
\[\frac{d}{dx}\int_a^{u(x)}\cos\left(e^t\right)~dt=\frac{d}{dx}\left[g(u(x))-g(a)\right]\]
Note that \(g(a)\) is a constant, so its derivative is 0:
\[\frac{d}{dx}\int_a^{u(x)}\cos\left(e^t\right)~dt=\frac{d}{dx}g(u(x))\]
By the chain rule, you have
\[\frac{d}{dx}g(u(x))=g'(u(x))\cdot u'(x)\]
Does that make sense?

Answer 13

yes, perfect, thanks a lot

Answer 14

You're welcome