# Finding Derivatives

1. Jan 3, 2006

### Jacobpm64

Find dy/dx if y = 4^(-x + 3)

I really don't know how to go about doing this.. never had to take the derivative of something with a variable as an exponent..

Thanks for the help in advance.

2. Jan 3, 2006

### Tx

This derivative requires the logarithmic change of base rule.
log_ax = lnx/lna
So this would become:
F(x) = 4^(3-x)
F'(x) = ln4 * (-4^(3-x))
= -2ln2 * 4^(3-x)

3. Jan 3, 2006

### TD

The derivative of $a^x$ is $\ln{a} \cdot a^x$ but don't forget the chain rule.

Last edited: Jan 3, 2006
4. Jan 3, 2006

### Little_Rascal

You probably meant that the derivative of $a^x$ is $\ln{a} \cdot a^x$?

5. Jan 3, 2006

### TD

Of course (it's still morning here )

Adjusted.

6. Jan 3, 2006

### HallsofIvy

Staff Emeritus
Or just: if y= ax, then ln(y)= x ln(a). Differentiate both sides, using the chain rule on the left.

7. Jan 3, 2006

### Jacobpm64

ok i understand that the formula for derivative of a^x is ln a * a^x

so i understand the step
F'(x) = ln 4 * (-4^(3-x)) <--- how did the 4 become negative?
and then i don't understand the next step at all...

8. Jan 3, 2006

### TD

The 4 becomes -4 because of the chain rule, since it says (3-x) and not just x, you have to multiply with the derivative of (3-x), which is -1 causing the sign change.
Then he just used a property of logaritms, being that ln(a^b) = b.ln(a) so ln(4) = ln(2²) = 2ln(2).

9. Jan 3, 2006

### Jacobpm64

ah, thanks a lot :)

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