Derivative of natural log function questions

In summary, the conversation discusses a student's difficulties with solving derivative of natural logarithm functions. They provide several equations and attempt to find the solutions using the chain rule and product rule. Some of their attempts are incorrect and they are seeking help to correct their mistakes. They also mention that these types of problems may appear on a midterm or final exam.
  • #1
SYoungblood
64
1

Homework Statement

Hello, I had a few derivative of the natural logarithm functions questions. It seems like it should be fairly straightforward, but I am turning it into a pig’s ear.On my honor, none of these are problems on an assessment per se, however, they are not materially different than anything that I am finding in my calc book and what I would assume to be on a midterm or final:

Homework Equations


[/B]
a) ln 3x

b) ln 3x + x

c) ln (3x + 4)

d) ln (x^2)

e) ln x^2

f) x (ln 2x)^4

The Attempt at a Solution



a) I believe I got this right – 1/3x

b) I think I have this one as well – 1/3x + 1

c) I have (3x + 4)’/(3x + 4) = 3/(3x + 4)

d) Using the chain rule, I believe this is 2x/x^2

e) I believe using the ln x derivative rules, this is 2 (ln x)’ = 2/x

f) This is giving me problems. Using the product rule and chain rule, I think the answer

= 1 * (ln 2x)^4 + x * 4(ln 2x)^3 * 1/2x

= (ln2x)^4 + 4 [(ln 2x)^3]/x, with an x in the numerator and denominator in the final term cancelling each other out.

Thank you for your help --

SY
 
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  • #2
SYoungblood said:

Homework Statement

Hello, I had a few derivative of the natural logarithm functions questions. It seems like it should be fairly straightforward, but I am turning it into a pig’s ear.On my honor, none of these are problems on an assessment per se, however, they are not materially different than anything that I am finding in my calc book and what I would assume to be on a midterm or final:

Homework Equations


[/B]
a) ln 3x

b) ln 3x + x

c) ln (3x + 4)

d) ln (x^2)

e) ln x^2

f) x (ln 2x)^4

The Attempt at a Solution



a) I believe I got this right – 1/3x
Maybe, but you didn't write it correctly. I think you mean 1/(3x) but what you wrote is (1/3)x.

Also, you need to use the chain rule.
SYoungblood said:
b) I think I have this one as well – 1/3x + 1
As above. This should be written as 1/(3x) + 1. You need to use the chain rule, though.
SYoungblood said:
c) I have (3x + 4)’/(3x + 4) = 3/(3x + 4)
Looks good
SYoungblood said:
d) Using the chain rule, I believe this is 2x/x^2
This should be simplified. Since ln(x2 = 2ln(x), its derivative is 2/x.
SYoungblood said:
e) I believe using the ln x derivative rules, this is 2 (ln x)’ = 2/x
Aren't d and e the same question?
That is, unless you mean (ln(x))2
SYoungblood said:
f) This is giving me problems. Using the product rule and chain rule, I think the answer

= 1 * (ln 2x)^4 + x * 4(ln 2x)^3 * 1/2x

= (ln2x)^4 + 4 [(ln 2x)^3]/x, with an x in the numerator and denominator in the final term cancelling each other out.
Did you forget to write the factor of x in the numerator?
I agree that the x factors in the numerator and denominator cancel.
SYoungblood said:
Thank you for your help --

SY
 
Last edited:
  • #3
SYoungblood said:
a) ln 3x

b) ln 3x + x

Is that what you meant or do you mean ln(3x+x)? Just curious, could be either way.

The Attempt at a Solution



a) I believe I got this right – 1/3x
Nope. Needs chain rule, even if you meant 1/(3x).
b) I think I have this one as well – 1/3x + 1
Nope Same problem.
 
  • #4
a) I forgot to write it as 1/ (3x) -- it looked better handwritten on paper.

b) Same traffic -- it should say [1/(3x)] + 1

c) Woo-hoo! -- basically a thru c are good to go.

d) I have that written as ln (x^2) -- that is where I had the chain rule as giving a derivative of 2x/(x^2) = 2/x

e) That one should be (ln x)^2 -- 2 ln x, again with an answer of 2/x

f) y = x(ln 2x)^4

Not sure if I missed an x in the first attempt, but here, I still have the product rule and chain rule as:

y' = 1 * (ln 2x)^4 + x * 4(ln 2x) ^3 * 1/(2x)

y' = (ln 2x) ^4 + 4x (ln 2x) ^3 / (2x) -- the 4x in the numerator and 2x in the denominator cancel out, with a numerator of 2 left

y' = (ln 2x)^4 + 2 (ln 2x)^3

Thoughts?

Thanks for all help

SY
 
  • #5
SYoungblood said:
a) I forgot to write it as 1/ (3x) -- it looked better handwritten on paper.

b) Same traffic -- it should say [1/(3x)] + 1

c) Woo-hoo! -- basically a thru c are good to go.

d) I have that written as ln (x^2) -- that is where I had the chain rule as giving a derivative of 2x/(x^2) = 2/x

e) That one should be (ln x)^2 -- 2 ln x, again with an answer of 2/x

f) y = x(ln 2x)^4

Not sure if I missed an x in the first attempt, but here, I still have the product rule and chain rule as:

y' = 1 * (ln 2x)^4 + x * 4(ln 2x) ^3 * 1/(2x)

y' = (ln 2x) ^4 + 4x (ln 2x) ^3 / (2x) -- the 4x in the numerator and 2x in the denominator cancel out, with a numerator of 2 left

y' = (ln 2x)^4 + 2 (ln 2x)^3

Thoughts?

Thanks for all help

SY

Let's just start with the first one. As LCKurtz has already pointed out, the derivative is not 1/(3x). Fix that one first.
 
  • #6
SYoungblood said:
...

e) That one should be (ln x)^2 -- 2 ln x, again with an answer of 2/x

f) y = x(ln 2x)^4

Not sure if I missed an x in the first attempt, but here, I still have the product rule and chain rule as:

y' = 1 * (ln 2x)^4 + x * 4(ln 2x) ^3 * 1/(2x)

y' = (ln 2x) ^4 + 4x (ln 2x) ^3 / (2x) -- the 4x in the numerator and 2x in the denominator cancel out, with a numerator of 2 left

y' = (ln 2x)^4 + 2 (ln 2x)^3

Thoughts?

Thanks for all help

SY
f) now looks good.

- but (e) is another matter.
## \left(\ln(x)\right)^2\ne 2\ln(x)##

Use the chain rule. This is much like the ##\ \left(\ln(2x)\right)^4\ ## portion of part (f), but simpler.
 
  • #7
SammyS said:
f) now looks good.

- but (e) is another matter.
## \left(\ln(x)\right)^2\ne 2\ln(x)##

Use the chain rule. This is much like the ##\ \left(\ln(2x)\right)^4\ ## portion of part (f), but simpler.

y = (ln x)^2

y' = 2 ln x * 1/x

y' = (2 ln x)/x

I think I got it this time.

Also, another above problem --

y = ln 3x

y' = 1/(3x) -- correct or not?

SY
 
  • #8
SYoungblood said:
y = (ln x)^2

y' = 2 ln x * 1/x

y' = (2 ln x)/x

I think I got it this time.

Also, another above problem --

y = ln 3x

y' = 1/(3x) -- correct or not?

SY

First one correct, second one not correct. Try the chain rule.
 
  • #9
Dick said:
First one correct, second one not correct. Try the chain rule.

y' = [1/(3x)] * 3 = 3/(3x) = 1/x
 
  • #10
SYoungblood said:
y' = [1/(3x)] * 3 = 3/(3x) = 1/x

Now it's right. You could also write ln(3x)=ln(3)+ln(x). Now it's pretty obvious the derivative is just 1/x.
 
  • #11
Thanks to all for the help. Derivatives of ln x and I had a long day today.
 

Related to Derivative of natural log function questions

What is the derivative of ln(x)?

The derivative of ln(x) is equal to 1/x.

What is the derivative of ln(u) where u is a function of x?

The derivative of ln(u) is equal to 1/u multiplied by the derivative of u with respect to x.

Can the derivative of ln(x) be negative?

Yes, the derivative of ln(x) can be negative for values of x less than 1.

Is the derivative of ln(x) the same as the derivative of log(x)?

Yes, the derivative of ln(x) and log(x) are the same. The notation "ln" is used for the natural logarithm and "log" is used for the base 10 logarithm.

How can I use the derivative of ln(x) in real-world applications?

The derivative of ln(x) can be used in various fields of science and engineering, such as physics, chemistry, and economics. It is particularly useful in calculating rates of change or growth, as well as determining maximum and minimum values of a function.

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