Solve Derivative of y=e^cosx*sinx

[Plutonium88]

Homework Statement

Find the derivativce of y=e^cosx*sinx

Homework Equations

3. The Attempt at a Solution

i`m just unsure of whether e should be treated as e^x because its not that.. or if e should be treated like an f(x)a^x to f`(x)a^xlna, or if it is still considerd e^x.. here is my attempt.

y`=e^cosx*(-sinx)*(sinx) + (cosx)(e^cosx)

y`= e^cosx(cosx - sin^2x)

 

[SammyS]

Homework Statement

Find the derivative of y=e^cosx*sinx

Homework Equations

3. The Attempt at a Solution

i`m just unsure of whether e should be treated as e^x because its not that.. or if e should be treated like an f(x)a^x to f`(x)a^xlna, or if it is still considered e^x.. here is my attempt.

y`=e^cosx*(-sinx)*(sinx) + (cosx)(e^cosx)

y`= e^cosx(cosx - sin^2x)

If you're differentiating y = (sin(x))e^cos(x), then what you have done is fine.

Writing a^x as e^{x ln(a)} can be helpful at times, but writing e in terms of some other constant generally isn't helpful when differentiating.

 

[Plutonium88]

If you're differentiating y = (sin(x))e^cos(x), then what you have done is fine.

Writing a^x as e^{x ln(a)} can be helpful at times, but writing e in terms of some other constant generally isn't helpful when differentiating.

Ah i just realized something... e^x = e^cosx in the sense that cosx is `x`. right?

And also thank you for your help.

 

[Fredrik]

Homework Equations

The chain rule. $(f\circ g)'(x)=f'(g(x))g'(x)$
The product rule. $(fg)'(x)=f'(x)g(x)+f(x)g'(x)$

i`m just unsure of whether e should be treated as e^x because its not that..

Keep in mind that $e^{f(x)}=exp(f(x))=(\exp\circ f)(x)$. So you will need the chain rule. You will also need the product rule to find f'(x).

 

[Plutonium88]

The chain rule. $(f\circ g)'(x)=f'(g(x))g'(x)$
The product rule. $(fg)'(x)=f'(x)g(x)+f(x)g'(x)$



Keep in mind that $e^{f(x)}=exp(f(x))=(\exp\circ f)(x)$. So you will need the chain rule. You will also need the product rule to find f'(x).

all right, thanks man.