Derivative of Exponential and Logarithmic Functions

Click For Summary
The discussion revolves around finding the equation of the tangent line to the graph of y = ln(x^2) at the point (2, ln4). The derivative is calculated as y' = 2/x, leading to a tangent slope of 1 at the specified point. There is a clarification regarding the notation of the logarithmic function, with suggestions that the original problem may contain a typo. Participants emphasize the importance of using the constant multiple rule instead of the product rule for derivatives involving constants to avoid unnecessary complexity. Overall, the conversation highlights efficient differentiation techniques and clarifies potential misunderstandings in the problem statement.
domyy
Messages
196
Reaction score
0

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
 
Physics news on Phys.org
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! :)
 

Similar threads

Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 40 ·
2
Replies
40
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
26
Views
3K
Replies
1
Views
1K