High School Derivation with logarithms and product

[Elias Waranoi]

I went through an example question that showed me how to solve the question but I'm not sure if I've misunderstood something or if they did a mistake.

Question: Derivate y = (1/a^x)^ax

ln(y) = ln( (1/a^x)^ax ) = ax( ln(1) - ln(a^x) ) = -ax ln(a^x)
(1/y)(dy/dx) = -ax * a^x ln(a) - a * ln(a^x)
dy/dx = (1/a^x)^ax * (-ax * a^x ln(a) - a * ln(a^x))

I think that they are derivating -ax ln(a^x) as a product of two functions with u*(dv/dx) + v*(du/dx) where u = -ax and v = ln(a^x). But isn't (dv/dx) supposed to be only ln(a)? So dy/dx = (1/a^x)^ax * (-ax * ln(a) - a * ln(a^x))

 

[stevendaryl]

I went through an example question that showed me how to solve the question but I'm not sure if I've misunderstood something or if they did a mistake.

Question: Derivate y = (1/a^x)^ax

ln(y) = ln( (1/a^x)^ax ) = ax( ln(1) - ln(a^x) ) = -ax ln(a^x)
(1/y)(dy/dx) = -ax * a^x ln(a) - a * ln(a^x)
dy/dx = (1/a^x)^ax * (-ax * a^x ln(a) - a * ln(a^x))

I think that they are derivating -ax ln(a^x) as a product of two functions with u*(dv/dx) + v*(du/dx) where u = -ax and v = ln(a^x). But isn't (dv/dx) supposed to be only ln(a)? So dy/dx = (1/a^x)^ax * (-ax * ln(a) - a * ln(a^x))

Your derivation simplifies a lot if you use the fact that ln(a^x) = x ln(a)

 

[fresh_42]

I went through an example question that showed me how to solve the question but I'm not sure if I've misunderstood something or if they did a mistake.

Question: Derivate y = (1/a^x)^ax

ln(y) = ln( (1/a^x)^ax ) = ax( ln(1) - ln(a^x) ) = -ax ln(a^x)
(1/y)(dy/dx) = -ax * a^x ln(a) - a * ln(a^x)
dy/dx = (1/a^x)^ax * (-ax * a^x ln(a) - a * ln(a^x))

I think that they are derivating -ax ln(a^x) as a product of two functions with u*(dv/dx) + v*(du/dx) where u = -ax and v = ln(a^x).

Yes.

But isn't (dv/dx) supposed to be only ln(a)? So dy/dx = (1/a^x)^ax * (-ax * ln(a) - a * ln(a^x))

No.

Try to differentiate $\ln a^x$ step by step.
What is $\frac{d}{dx}\ln f(x)$ and how do you differentiate $f(x)=a^x$?

Hint: You can write $f(x)=a^x=\exp(\ln a^x)=\exp(x\ln a)$.

 

[Elias Waranoi]

But isn't v = ln(a^x) = x ln(a)
dv/dx = ln(a)

I can get that dv/dx is a^x ln(a) if v = a^x so:
v = a^x -> ln(v) = ln(a^x) = x ln(a)
(1/v)(dv/dx) = ln(a) -> dv/dx = v ln(a) = a^x ln(a)

But v = ln(a^x) so I can't get anything but ln(a)

 

[fresh_42]

But isn't v = ln(a^x) = x ln(a)
dv/dx = ln(a)

I can get that dv/dx is a^x ln(a) if v = a^x so:
v = a^x -> ln(v) = ln(a^x) = x ln(a)
(1/v)(dv/dx) = ln(a) -> dv/dx = v ln(a) = a^x ln(a)

But v = ln(a^x) so I can't get anything but ln(a)

You're right. I first mistakenly forgot a denominator. Maybe the author you have your example from made the same mistake.

 

[Mark44]

Question: Derivate y = (1/a^x)^ax

Although "derivate" is a word in English, it is not used in mathematics. To obtain the derivative of a function, you differentiate it or find its derivative.