Is My Derivative Calculation for a Logarithmic Function Correct?

In summary: The latter is correct, unfortunately for you :)You're not completely off, though, as \frac{2x - 3}{x^2 - 1} = \frac{3 x - 4}{x^2 - 1} + \frac{x - 1}{x^2 - 1} = \frac{3 x - 4}{x^2 - 1} - \frac{1}{1 + x}so it looks like you're just missing a term or you've got a sign wrong.Also unfortunately, it is hard for us to tell you what went wrong without showing us your work. It's basically just calculating: d/du log(u) = 1/u,
  • #1
xxclaymanxx
7
0
1. Given Y = ln [ (x+1)^3/((x^2)-1)^(1/2), find y'



2. I came out with the following answer to this question:

(3x-4)/((x^2)-1)


How ever, I typed the question into an online derivative calculator (to hopefully check my asnwer as I have no answer key, and want to make sure I'm on the right path), but it came up with a completely different answer:

(2x-3)/((x^2)-1)

Could anyone point me in the right direction...my answer worked out nicely: factored, canceled etc. but I'm worried its not correct.

Thanks for the check!
 
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  • #2
The latter is correct, unfortunately for you :)
You're not completely off, though, as
[tex]\frac{2x - 3}{x^2 - 1} = \frac{3 x - 4}{x^2 - 1} + \frac{x - 1}{x^2 - 1} = \frac{3 x - 4}{x^2 - 1} - \frac{1}{1 + x}
[/tex]
so it looks like you're just missing a term or you've got a sign wrong.

Also unfortunately, it is hard for us to tell you what went wrong without showing us your work. It's basically just calculating: d/du log(u) = 1/u, using the chain rule with u = (x+1)^3/((x^2)-1)^(1/2).
 
  • #3
Thank you for your help! it allowed me to go back into my work, and figure out where I wen't wrong. Basically all I did, was i forgot to write an X, and instead wrote a 1...so when I was multiplying both sides by a common demonator, my numbers came out funny.

Anyways, I found the error, corrected the following calculations, and VOILA! got it.

Thanks again!
 
  • #4
xxclaymanxx said:
1. Given Y = ln [ (x+1)^3/((x^2)-1)^(1/2)], find y'

May I point out that Y= 3ln(x+1)-(1/2)ln(x2+ 1). Surely that is simpler to differentiate!



2. I came out with the following answer to this question:

(3x-4)/((x^2)-1)


How ever, I typed the question into an online derivative calculator (to hopefully check my asnwer as I have no answer key, and want to make sure I'm on the right path), but it came up with a completely different answer:

(2x-3)/((x^2)-1)

Could anyone point me in the right direction...my answer worked out nicely: factored, canceled etc. but I'm worried its not correct.

Thanks for the check!
 

FAQ: Is My Derivative Calculation for a Logarithmic Function Correct?

What is the derivative of a log function?

The derivative of a log function, also known as the logarithmic function, is the inverse of the natural logarithm function. It is denoted as ln(x) or loge(x) and is equal to 1/x.

How do you find the derivative of a log function?

To find the derivative of a log function, you can use the logarithmic differentiation method. This involves taking the natural log of both sides of the function, using logarithmic rules to simplify, and then applying the power rule or chain rule to find the derivative.

What is the derivative of log base a of x?

The derivative of log base a of x, or loga(x), is equal to 1/(xln(a)). This can be derived by using the change of base formula and the chain rule.

Can you provide an example of finding the derivative of a log function?

Sure, let's find the derivative of y = ln(3x). Using the logarithmic differentiation method, we get ln(y) = ln(3x). Applying the power rule, we get 1/y * dy/dx = 3/x. Solving for dy/dx, we get dy/dx = 3/(x * y). Substituting y = ln(3x), we get dy/dx = 3/(x * ln(3x)).

What is the significance of the derivative of a log function?

The derivative of a log function is important in many mathematical and scientific applications. It is commonly used in calculus to find the slope of a curve, the rate of change of a function, and to solve optimization problems. In physics and engineering, it is used to model various natural phenomena and in finance, it is used in compound interest calculations and financial modeling.

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