Derivative of Exponential and Logarithmic Functions

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Homework Help Overview

The discussion revolves around finding the equation of the tangent line to the graph of the function y = ln(x^2) at the point (2, ln(4)). Participants explore the differentiation of logarithmic functions and the application of derivative rules.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the differentiation process, with some attempting to clarify the use of the product rule versus the constant multiple rule. Questions arise regarding the interpretation of the function, specifically whether it is ln(x^2) or (ln(x))^2.

Discussion Status

There is an ongoing exploration of the correct application of differentiation rules, with some participants providing hints and clarifications. Multiple interpretations of the function are being considered, and guidance on simplifying the differentiation process has been offered.

Contextual Notes

Some participants express confusion regarding a potential typo in the answer sheet they received, which states the answer should be y = x - 2 + ln without specifying the argument of the logarithm. This raises questions about the accuracy of the provided answer.

domyy
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Homework Statement



Find the equation of the tangent line to the graph of y = lnx2 at the point (2, ln4)

Homework Equations


The Attempt at a Solution



y' = (lnx2)'
y' = 2lnx
y' = (2)(lnx)' + (lnx)(2)'
y' = 2/x

Mtan/(2,ln4) = 2/2 = 1

Equation of tan line:
y - ln4 = 1(x-2)
y = x - 2 + ln4
 
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domyy said:

Homework Statement



Find the equation of the tangent line to the graph of y = lnx2 at the point (2, ln4)

Homework Equations





The Attempt at a Solution



y' = (lnx2)'
y' = 2lnx
y' = (2)(lnx)' + (lnx)(2)'
y' = 2/x

Mtan/(2,ln4) = 2/2 = 1

Equation of tan line:
y - ln4 = 1(x-2)
y = x - 2 + ln4

You forgot a prime in the second line, but that looks fine.
 
I was looking at the answer sheet I was given along with the exercise and it says the answer should be y = x - 2 + ln

Is it a typo ? Or I am wrong?
 
Thanks for replying! :)
 
domyy said:
I was looking at the answer sheet I was given along with the exercise and it says the answer should be y = x - 2 + ln

Is it a typo ? Or I am wrong?

If that's what the answer really says, then they omitted the argument of the log. Sure it's a typo.
 
Hint: the derivative of a constant times a function is the constant times the derivative of the function:

y = c * f(x)
y' = c * f'(x)

The use of the product rule in these instances is not necessary.
 
You have y= ln x^2. Is that y= ln(x^2) or y= (ln(x))^2? If it is the first, as you seem to have assumed, then it is easier to write it as y= 2 ln(x) so that y'= 2/x immediately.
 
Oh thanks!
The observations helped a lot. I was taking much longer writing down the whole process to find the derivatives. These tips are saving me a lot of time. THANK YOU SO MUCH!
 
Last edited:
domyy said:
]

y' = (lnx2)'
y' = 2lnx
y' = (2)(lnx)' + (lnx)(2)'
y' = 2/x

SteamKing said:
Hint: the derivative of a constant times a function is the constant times the derivative of the function:

y = c * f(x)
y' = c * f'(x)

The use of the product rule in these instances is not necessary.

I recommend that students NEVER use the product rule when one factor of a product is a constant, as in the problem above. What SteamKing is describing is often called the constant multiple rule, which says d/dx( k *f(x)) = k * d/dx(f(x)).

Also, you should NEVER use the quotient rule if the denominator is a constant, such as x3/6. Write this as (1/6) * x3 and use the constant multiple rule.

The reason for these recommendations is that the product rule is more complicated than the constant multiple rule, so you are more likely to make mistakes. And the quotient rule is even more complicated, increasing your chances of making an error.
 
  • #10
I just solved one problem using this technique. Oh I am loving these tips :) they save time! :)
 

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