Question

what the derivative for the function y=2.2^x ? if you can plzz

Answer 1

y'=x2.2^x-1

Answer 2

\[y = 2.2^x \Rightarrow \ln(y) = \ln(2.2^x) \Rightarrow \ln(y) = xln(2.2)\]
Taking the derivative of both sides we have:
\[\frac{1}{y}y' = \ln(2.2) \Rightarrow y' = yln(2.2)\]
but:
\[y = 2.2^x\]
so plugging that in we obtain our answer:
\[y' = \ln(2.2)*2.2^x\]

Answer 3

hey why do you put ln you can easily solve it using the basic differentiation rules

Answer 4

He is explaining that y = a^x means x= log_a y.

Answer 5

@hashir your answer is wrong. you cant treat:
\[y = a^x\]
like:
\[y = x^n\]
They are totally different.