Derivative of natural log function questions

[SYoungblood]

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

 

[Mark44]

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.

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.

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

Looks good

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.

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

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.

Thank you for your help --

SY

 

[LCKurtz]

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.

 

[SYoungblood]

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

 

[Dick]

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.

 

[SammyS]

...

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.

 

[SYoungblood]

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

 

[Dick]

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.

 

[SYoungblood]

First one correct, second one not correct. Try the chain rule.

y' = [1/(3x)] * 3 = 3/(3x) = 1/x

 

[Dick]

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.

 

[SYoungblood]

Thanks to all for the help. Derivatives of ln x and I had a long day today.