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Derivative of e^sinx with Chain Rule | Calculus 1 Exercises
Wrath of Math
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We differentiate e^sinx using the chain rule. The outside function f(x) is f(x) = e^x, and the inside function g(x) is g(x)=sinx. Then, f(g(x)) = e^sinx, so we see we've identified the outside and inside functions correctly. Applying the chain rule, we have the derivative of f(g(x)) is f'(g(x))*g'(x) = e^(sinx)*(cosx). #calculus1 #apcalculus
Calculus 1 Exercises playlist: https://www.youtube.com/playlist?list=PLztBpqftvzxUEqGGgvL3EuIQUNcAdmVhx
Calculus 1 playlist: https://www.youtube.com/playlist?list=PLztBpqftvzxWVDpl8oaz_Co6CW50KtGJy
Chain Rule Explained: https://youtu.be/vpVJoFEY_xM
Derivative of e^2x with Chain Rule: https://youtu.be/3b0SyiCD2eo
Derivative of e^3x with Chain Rule: https://youtu.be/CFY80JrLSR0
Derivative of e^(1/x): https://youtu.be/PtmGHEJP_Bk
Derivative of e^(cosx): https://youtu.be/9MB06MCxp74
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@@ruffstuffOG
Thanks a lot sir
@@smaransuthar663
Thank you