How can i find the derivatives of a^x ?

Question

i tried to find the derivativesof a^x in different ways.But later i heard that it can be solved by using binomial theorem.how and why binomial theorem will be used to find the derivatives of a^x?

Answer 1

Actually, a^x = exp(x*ln(a)), so d(a^x) /dx = ln(a)*a^x.

Easy. (Oh, and I used substitution.)

No need for binomial thm.

Answer 2

David makes a basic mistake. His formula would work if the x was not in the exponent, but was only the base. Then the derivative of x^a is
a x^(a-1).
But for situations where x is in the exponent, you have to write
a^x=exp(x*(ln a)), so
the derivative will be
a^x (ln a).

So im2 is correct.

Answer 3

The spinoff of y = e^x is y´= e^x however the spinoff of y = e^(ax) we ought to apply the chain rule. y´= a.e^(ax) The reduce (x->0) of ((a million-e^ax)/(x^7)) is the reduce (x->0) of (-a.e^ax) /(7x^6 ) ... right here the numerator has a tendency to -a and the denominator has a tendency to 0. then the reduce is infinity... The reduce does no longer exist because the image grows to infinity. ok!

Answer 4

Binomial theorem is not a good choice. take log

f = a ^ x
log f = x log a

df/dx(1/f) = log a or df/dx = f log a = log a(a^x)

Answer 5

y = a^x

taking log on both the sides we have logy = loga^x = xloga
differentiating both sides, (1/y)y' = loga
y' = yloga

y' = a^xloga

Answer 6

im2_weird has the correct answer.

d(a^x) / dx = ln(a) * a^x

Answer 7

im2_weird is correct.

Answer 8

d(a)/d(x) of a^x is xa^(x-1)