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f(x)= 2^e^sinx find derivative
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@Walker21 Hi :) \(\huge \color{red}{\text{Welcome to Open Study}}\ddot\smile\) do you know the chain rule ?? also, first you can put 2^e = a (another constant) here....
Did Goku just use instant transmission?
lol nvm
so chain rile first?
you'll need chain rule for a^(sin x) because standard function is a^x
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ok
still lost
Take e^sinx=u d(2^u)/dx= 2^u log2 du/dx Sloving for du/dx = d(e^sinx)/dx = e^sinx (cosx) So final answer is 2^(e^sinx).log2.e^sinx .cosx
I hope it was clear ...
thanks helps
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Medal is its accurate :)
are you good at finding points on a graph at which the tagent line is horizonatal?
Not so much. No
ok
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