Question

Question about n^x and the derivative. So I know that if f(x)=n^x, then f'(x)=n^x*ln(n). I don't really understand WHY though? I know the PROOFS, but I don't have a physical understanding as to WHY this is true. So if you have a number n raised to the power of x, the slope of the tangent line at the point n^x is y value of n^x, multiplied by the exponent required for base e to equal n? That's what I'm not understanding.

Answer 1

let y = n^x

we can take the log of each side

log(y) = log(n^x)

using log properties we get

log(y) = x log(n)

taking the derivative of each side we get

y'/y = log(n)

solve for y'

y' = y log(n)

but we know from the start that y = n^x, therefore

y' = n^x log(n)

Answer 2

really depends on your definition of \(f(x)=b^x\)

one is \(b^x=e^{x\ln(b)}\) so for example \(f(x)=2^x=e^{x\ln(2)}\) then take the derivative and you will see it right away

@amistre64 method also explains it

Answer 3

if n=e that reduces us to the (e^x)' = e^x log(e); or simply (e^x)' = e^x

Answer 4

the definition
\[b^x=e^{x\ln(b)}\] makes sense out of things like \(2^{\pi}\) for example, it becomes
\[e^{\pi\ln(2)}\]

it also allows you to make sense out of functions that have a variable in the base and the exponent.
for example
\[x^{\sin(x)}=e^{\sin(x)\ln(x)}\]

Answer 5

real analysis defines:\[log(x)=\int_{1}^{\infty}\frac1xdx\]
earlier math classes use a variety of log functions differentiated by bases ... but since they can all be configured in log(x) form, they simply define a "log" function. such that log(e) = 1

Answer 6

I understand the algebra behind it, I've seen many proofs. I'm trying to understand how the slope is related to "the exponent that e must be raised by to equal n". I understand how n^x can affect a slope, that's very obvious. But why is ln(n) altering the slope? What's that relation?

Answer 7

most likely your trying to consider how the chain rule works.

consider a box full of gears such that you can turn the first gear at a certain rate and it effects the last gears output rate. a transmission in a car if you will.

Answer 8

e^u -> u' e^u ; but then we have to ask ourselves, what is effecting u?

let u = x ln(n) ; u' = ln(n) for suitable constant n