I am supposed to find the nth derivative of 2^x. I am not sure where to start because with each time the derivative gets more complicated. Thanks in advance.
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f '(x) = 2^x * ln 2
f ''(x) = (2^x * ln 2) * ln 2 = 2^x * (ln 2)^2
f '''(x) = (2^x * ln 2) * (ln 2)^2 = 2^x * (ln 2)^3.
So, we see that
f^(n)(x) = 2^x * (ln 2)^n.
I hope this helps!
f ''(x) = (2^x * ln 2) * ln 2 = 2^x * (ln 2)^2
f '''(x) = (2^x * ln 2) * (ln 2)^2 = 2^x * (ln 2)^3.
So, we see that
f^(n)(x) = 2^x * (ln 2)^n.
I hope this helps!