# Derivative of an exponential function

1. Jan 24, 2009

### magma_saber

1. The problem statement, all variables and given/known data
What is the derivative of ex3? also what is the derivative of (ln1/x)2

2. Relevant equations

3. The attempt at a solution
is it 3x2e3x2?
2(ln1/x)(x)?

Last edited: Jan 24, 2009
2. Jan 24, 2009

### jgens

Re: Derivative

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.

3. Jan 24, 2009

### Staff: Mentor

Re: Derivative

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.

Last edited: Jan 24, 2009
4. Jan 24, 2009

### magma_saber

Re: Derivative

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

5. Jan 24, 2009

### Staff: Mentor

Re: Derivative

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

6. Jan 24, 2009

### jgens

Re: Derivative

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.

7. Jan 24, 2009

### jgens

Re: Derivative

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

8. Jan 24, 2009

### magma_saber

Re: Derivative

so its 3x2ex3?

9. Jan 24, 2009

### jgens

Re: Derivative

Correct.

10. Jan 24, 2009

### magma_saber

Re: Derivative

d(ln1/x) is x^2?

11. Jan 24, 2009

### jgens

Re: Derivative

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

12. Jan 24, 2009

### The Dagda

Re: Derivative

Might be easier if you just look at it as:

$$\frac{d}{dx}\;(-ln(x))$$

Then it's just straight forwardly obvious.

13. Jan 24, 2009

### jgens

Re: Derivative

That certainly should make it simpler for the original poster!