Question

Derive cos(e^x+1)

I gte the answer is -sin(e^x) but it is wrong it says the answer is -e^x(e^x+1)

Answer 1

Do you mean "take the derivative"?

To derive and to take the derivative are two different things...

I'm suspicious that you haven't copied the problem and/or answer correctly.

Answer 2

\(\bf cos(e^{x+1})?\qquad cos(e^x+1)?\)

Answer 3

yea the second one and its take the derivative ca nyou help

Answer 4

I think it is likely to be the latter, and the correct answer there would be \[-e^x\sin(e^x+1)\]

Answer 5

yes but how do I get that? I get -sin(e^x)

Answer 6

It's just the chain rule. let \[u = e^x+1\]then you have \[\frac{du}{dx} = e^x\]and your problem becomes \[\frac{d}{dx}[\cos(e^x+1)] = \frac{d}{du}[\cos(u)] *\frac{du}{dx} = \]

Answer 7

what is \[\frac{d}{du}[\cos(u)]=\]

Answer 8

that's just \[-\sin(u)\]isn't it? What do you get after you reverse the substitution of \[u = e^x+1\]? Then multiply by the value of \[\frac{du}{dx}\]

Answer 9

oh yeah the chain rule is the derivativre multiplied by original equation right?

Answer 10

of u

Answer 11

you have a function of a function here. if we call the outer function f, and the inner function g, then we have \[f(g(x)) = \cos(e^x+1)\]where \[g(x) = e^x+1\]

To find the derivative, we find the derivative of f(g(x)) and multiply it by the derivative of g(x). We did that by letting (u = g(x)\) and finding the derivative of f(u), then undoing the substitution.

Answer 12

For example, to find the derivative of \[e^{3x}\]we let \(u = 3x\) and then \(du = 3\,dx\) and our problem becomes \[\frac{d}{dx}[e^{3x}] = \frac{d}{du}[e^u]*\frac{du}{dx} = e^u*e = e^{3x}*3 = 3e^{3x}\]

Or similarly to find \[\frac{d}{dx}[\sin(2x)] = \frac{d}{dx}[\sin(u=2x)] = \frac{d}{du}[\sin(u)]*\frac{d}{dx}[u] = \cos(u)*2 = 2\cos(2x)\]

Answer 13

Sorry, last bit got cut off:
\[ \cos(u)*2 = 2\cos(2x)\]