- #1

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Hello all. I would like to know what is the derivative of 2^x and how it is done. Thank you

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- Thread starter l33t_V
- Start date

- #1

- 8

- 0

Hello all. I would like to know what is the derivative of 2^x and how it is done. Thank you

- #2

uart

Science Advisor

- 2,776

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Here's a hint, start by expressing it as :

[tex] 2^x = e^{x \ln(2)}[/tex]

[tex] 2^x = e^{x \ln(2)}[/tex]

- #3

- 8

- 0

I'll give it a try. But why can't it be solved as we do x^k ? Like k*x'*x^(k-1)

- #4

- 811

- 6

I'll give it a try. But why can't it be solved as we do x^k ? Like k*x'*x^(k-1)

Because that's against the rules.

The rule is

[tex]\frac{d}{dx}(x^n) = n x^{n-1}[/tex]

The base has to be the variable being differentiated against. The exponent has to be constant with respect to the base.

Luckily, all is not lost. We have lots of rules in calculus. Instead, we can use this one.

[tex]\frac{d}{dx}(e^{f(x)}) = f'(x) e^{f(x)}[/tex]

This rule can only be used when the base is the Euler number e and when the exponent is a function of x. If we want to use this rule for your problem, we can do a little advanced algebra to change the base (which is what uart suggested).

Some problems in calculus can't be solved exactly, even if they look super simple. For example, if you have the function [tex]f(x) = x^x[/tex], you can use NEITHER rule.

- #5

lurflurf

Homework Helper

- 2,440

- 138

[tex]\frac{d}{dx}u^v=v u^{v-1} \frac{du}{dx}+\log(u) u^v \frac{dv}{dx}[/tex]

Another thread about it.

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