The derivative of the function y = e^(cos(x)) * sin(x) is calculated using the product and chain rules of differentiation. The correct derivative is y' = e^(cos(x)) * (cos(x) - sin^2(x)). The discussion emphasizes that e should not be treated as e^x but rather as e raised to the power of a function, specifically e^(cos(x)). The application of the chain rule and product rule is essential in deriving the solution accurately.
PREREQUISITESStudents studying calculus, particularly those focusing on differentiation techniques, as well as educators looking for examples of applying the product and chain rules in real-world problems.
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).