Derivative of b^x
b^x ln(b)
The derivative of b^x is given by:
d/dx (b^x) = ln(b) * b^x
where ln(b) is the natural logarithm of b.
We can derive this using the following steps:
Start with the definition of the derivative:
d/dx (b^x) = lim(h→0) [b^(x+h) – b^x]/h
Apply exponential rule (a^b * a^c = a^(b+c)) to the numerator:
d/dx (b^x) = lim(h→0) [b^x * b^h – b^x]/h
Factor out b^x from the numerator:
d/dx (b^x) = lim(h→0) [b^x * (b^h – 1)]/h
Apply limit definition of exponential function, as h approaches zero, (b^h – 1)/h approaches the natural logarithm of b:
d/dx (b^x) = b^x * ln(b) * lim(h→0) [(b^h – 1)/h]
Using L’Hopital’s rule, we can evaluate the limit in the parenthesis. Take the derivative of the numerator and denominator with respect to h:
d/dx (b^x) = b^x * ln(b) * lim(h→0) [ln(b) * b^h]
As h approaches zero, ln(b) * b^h approaches ln(b) * 1 = ln(b).
Therefore, we get
d/dx (b^x) = b^x * ln(b) * ln(b) = ln(b) * b^x
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