Derivation with logarithms and product

In summary: Also, the function in this question is actually y = (1/ax)^ax, not y = (1/ax)ax.In summary, the conversation discusses the derivation of the function y = (1/ax)^ax and questions the correct method of differentiating -ax ln(ax). It is noted that ln(a^x) simplifies to x ln(a) and the correct value of dv/dx is ax ln(a). The author may have made a mistake in their example.
  • #1
Elias Waranoi
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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/ax)ax

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

I think that they are derivating -ax ln(ax) as a product of two functions with u*(dv/dx) + v*(du/dx) where u = -ax and v = ln(ax). But isn't (dv/dx) supposed to be only ln(a)? So dy/dx = (1/ax)ax * (-ax * ln(a) - a * ln(ax))
 
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  • #2
Elias Waranoi said:
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/ax)ax

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

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

Your derivation simplifies a lot if you use the fact that [itex]ln(a^x) = x ln(a)[/itex]
 
  • #3
Elias Waranoi said:
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/ax)ax

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

I think that they are derivating -ax ln(ax) as a product of two functions with u*(dv/dx) + v*(du/dx) where u = -ax and v = ln(ax).
Yes.
But isn't (dv/dx) supposed to be only ln(a)? So dy/dx = (1/ax)ax * (-ax * ln(a) - a * ln(ax))
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)##.
 
  • #4
But isn't v = ln(ax) = x ln(a)
dv/dx = ln(a)

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

But v = ln(ax) so I can't get anything but ln(a)
 
  • #5
Elias Waranoi said:
But isn't v = ln(ax) = x ln(a)
dv/dx = ln(a)

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

But v = ln(ax) 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.
 
  • #6
Elias Waranoi said:
Question: Derivate y = (1/ax)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.
 
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1. What is the basic concept behind derivation with logarithms and product?

The basic concept is that the derivative of a product of two functions can be found by taking the derivative of each individual function and then adding them together. In the case of logarithms, the derivative of ln(x) is 1/x, so when taking the derivative of a product involving logarithms, each individual logarithmic function is replaced with its derivative.

2. How do you use the logarithmic derivative rule in practice?

In practice, the logarithmic derivative rule can be used to simplify complex functions before taking the derivative. By breaking down the function into smaller, simpler components and using the logarithmic derivative rule on each component, the overall derivative can be found more easily.

3. What is the difference between the logarithmic derivative rule and the product rule?

The logarithmic derivative rule is a specific case of the product rule, where one of the functions is a logarithm. The product rule states that the derivative of a product of two functions is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function. The logarithmic derivative rule simply replaces the logarithmic function with its derivative.

4. Can the logarithmic derivative rule be applied to any function involving logarithms?

Yes, the logarithmic derivative rule can be applied to any function that contains logarithms. However, it is important to note that the function must be written in the form of a product before the rule can be applied.

5. How does the logarithmic derivative rule relate to the chain rule?

The chain rule states that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. The logarithmic derivative rule is a special case of the chain rule, where the outer function is ln(x) and the inner function is another function. The rule can be used to simplify the derivative of a composite function involving a logarithm.

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