The discussion revolves around finding the derivative of the function y = e^(cos(x)) * sin(x). Participants are exploring the application of differentiation rules in the context of this expression.
Some participants have provided guidance on the use of the chain and product rules, noting their relevance to the problem. There is acknowledgment of the attempts made, but no consensus on the correctness of the approaches has been reached.
Participants are grappling with the definitions and assumptions surrounding the function's components, particularly regarding the treatment of e and its relationship to the derivative process.
Plutonium88 said: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)
Ah i just realized something... e^x = e^cosx in the sense that cosx is `x`. right?SammyS said:If you're differentiating y = (sin(x))ecos(x), then what you have done is fine.
Writing ax as ex ln(a) can be helpful at times, but writing e in terms of some other constant generally isn't helpful when differentiating.
The chain rule. ##(f\circ g)'(x)=f'(g(x))g'(x)##Plutonium88 said:Homework Equations
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 said:i`m just unsure of whether e should be treated as e^x because its not that..
Fredrik said: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).