Derivative of an exponential function

[magma_saber]

Homework Statement

What is the derivative of e^x3? also what is the derivative of (ln1/x)²

Homework Equations

The Attempt at a Solution

is it 3x²e^3x2?
2(ln1/x)(x)?

 

[jgens]

Your solution for d(e^(x^3))/dx is incorrect. Remember d(e^u)/dx = u'e^u where u is a function of x. For the other derivative, let u = ln(1/x) and then apply the chain rule.

 

[Mark44]

No. No.
Both problems require the chain rule.
For the second problem, d/dx((ln(1/x))^2) = 2* ln(1/x) * d/dx(1/x). You made a mistake in your derivative of 1/x.

 

[magma_saber]

so the second is 2(ln1/x)(-x^-2)?
and the first one is x^3e^x^3?

 

[Mark44]

Right for the second one, but I would write it as 2 ln(1/x) (-1/x^2) or even better as
[tex]2 ln(\frac{1}{x}) \frac{-1}{x^2}[/tex]
Not right for the first. As jgens said, d/dx(e^u) = du/dx * e^u (where u = x^3).

 

[jgens]

Your first derivative is still incorrect. d(e^u)/dx = u'e^u, not ue^u.
Your second derivative is also incorrect. Mark gave you an incorrect expression to differentiate. You need to find 2*ln(x)*d(ln(1/x))/dx which also requires the chain rule.

 

[jgens]

Right for the second one, but I would write it as 2 ln(1/x) (-1/x^2) or even better as
[tex]2 ln(\frac{1}{x}) \frac{-1}{x^2}[/tex]

Don't forget d(ln(1/x))/dx, which certainly is not -1/x^2!

 

[magma_saber]

so its 3x²e^x3?

 

[jgens]

Correct.

 

[magma_saber]

d(ln1/x) is x^2?

 

[jgens]

No, let u = 1/x then you need to find d(lnu)/dx which will be (1/u)(du/dx).

 

[The Dagda]

Might be easier if you just look at it as:

[tex]\frac{d}{dx}\;(-ln(x))[/tex]

Then it's just straight forwardly obvious.

 

[jgens]

That certainly should make it simpler for the original poster!