Finding Derivative of y = 4^(-x + 3)

[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.

 

[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)

 

[TD]

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

 

[Little_Rascal]

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

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

 

[TD]

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

Of course (it's still morning here )

Adjusted.

 

[HallsofIvy]

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

 

[Jacobpm64]

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)

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...

 

[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).

 

[Jacobpm64]

ah, thanks a lot :)