Here we have g(t)=e6t3t
So, differentiating both sides:-
g(t)=e6t3t ⇒g′(t)=e12te6t(3)−3t(6e6t) ⇒g′(t)=e12t3e6t−18te6t ⇒g′(t)=e6t3−18t
And also we have f(x)=exsin(ex)
So, differentiating both sides:-
f(x)=exsin(ex)⇒f′(x)=sin(ex)ex+excos(ex)ex⇒f′(x)=exsin(ex)+e2xcos(ex)
So, we have our derivatives of the two given functions.
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