Ln Derivatives

[Whateverchan]

Hi guys,

I am stuck on these two log derivative problems. For the first one, I tried using ln to solve it.
(7x ln sqrt(x))' = 7 ln sqrt(x) + 7x (1/sqrt(x)) (1/2) (x)^-1/2.

For the second one, I think I have to solve it with implicit diff.
y' = 2ln(3x) + 2ln(5y)
y' = 2(1/3x)(3) + 2(1/5y)(5)(y')
y' - 2/y(y') = 2/x
y' = 2/x : (1-2/y)



What did I do wrong and miss?

By the way, why doesn't my avatar show up?

 

[Jason76]

Hi guys,

I am stuck on these two log derivative problems. For the first one, I tried using ln to solve it.
(7x ln sqrt(x))' = 7 ln sqrt(x) + 7x (1/sqrt(x)) (1/2) (x)^-1/2.

For the second one, I think I have to solve it with implicit diff.
y' = 2ln(3x) + 2ln(5y)
y' = 2(1/3x)(3) + 2(1/5y)(5)(y')
y' - 2/y(y') = 2/x
y' = 2/x : (1-2/y)

View attachment 3329

What did I do wrong and miss?

By the way, why doesn't my avatar show up?

For the 1st one, rewrite the square root as a power of $\displaystyle \dfrac{1}{2}$. Next, use ln differentiation. Note: You would simply using the ln exponent rule twice before taking the derivative. That's what I think. Can anyone verify this?

$\displaystyle y = \sqrt{x^{7x}}$

$\displaystyle y = (x^{7x})^{1/2}$

$\displaystyle \ln[y] = \ln[(x^{7x})^{1/2}]$

$\displaystyle \ln[y] = \dfrac{1}{2}\ln[(x^{7x})]$

$\displaystyle \ln[y] = \dfrac{1}{2} (7x) \ln[x]$

$\displaystyle \ln[y] = \dfrac{7}{2} x \ln[x]$

$\displaystyle \dfrac{1}{y}(y') = [\ln x][\dfrac{7}{2}] + [\dfrac{7}{2} x][\dfrac{1}{x}]$ - Product Rule $\displaystyle g'(f) + f(g')$ given $\displaystyle (f)(g)$

$\displaystyle (y') = (y) [\ln x][\dfrac{7}{2}] + [\dfrac{7}{2} x][\dfrac{1}{x}]$

$\displaystyle (y') = ( \sqrt{x^{7x}}) [\ln x][\dfrac{7}{2}] + [\dfrac{7}{2} x][\dfrac{1}{x}]$


:?: Right so far?

 

[Jason76]

For the 2nd one, you would take the derivative of $\displaystyle \ln$ Next, you would do implicit differentiation on $\displaystyle du$ Finally you would make the $\displaystyle y'$ terms one (isolated on the left side of the equation), and then back-substitute. Remember $\displaystyle \dfrac{d}{dx} \ln u = \dfrac{1}{u}(du)$

$\displaystyle y = \ln(3x^{2} + 5y^{2})$

$\displaystyle y' = \dfrac{1}{u}(6x + 10yy')$

$\displaystyle y' - 10yy' = \dfrac{1}{u}(6x)$

$\displaystyle y'[1 - 10y] = \dfrac{1}{u}(6x)$

$\displaystyle y' = \dfrac{1}{u}(6x)\dfrac{1}{(1 - 10y)}$

$\displaystyle y' = \dfrac{1}{3x^{2} + 5y^{2}}(6x)\dfrac{1}{(1 - 10y)}$

$\displaystyle y' = \dfrac{6x}{(3x^{2} + 5y^{2})(1 - 10y)}$ Think this is right.

 

[stapel]

For the 1st one...

$\displaystyle y = \sqrt{x^{7x}}$

According to the screenshot, the power "7x" is on the radical, not inside the radical on the x.

 

[stapel]

I am stuck on these two log derivative problems. For the first one, I tried using ln to solve it.
(7x ln sqrt(x))' = 7 ln sqrt(x) + 7x (1/sqrt(x)) (1/2) (x)^-1/2.

How did you get from the above to what you'd entered as your answer?

For the second one, I think I have to solve it with implicit diff.

Yes.

y' = 2ln(3x) + 2ln(5y)

How did you get this?

Note: ln(a + b) does not equal ln(a) + ln(b)!

 

[Jason76]

According to the screenshot, the power "7x" is on the radical, not inside the radical on the x.

ok, my mistake. It was difficult to see it's location.

 

[Whateverchan]

For the 2nd one, you would take the derivative of $\displaystyle \ln$ Next, you would do implicit differentiation on $\displaystyle du$ Finally you would make the $\displaystyle y'$ terms one (isolated on the left side of the equation), and then back-substitute. Remember $\displaystyle \dfrac{d}{dx} \ln u = \dfrac{1}{u}(du)$

$\displaystyle y = \ln(3x^{2} + 5y^{2})$

$\displaystyle y' = \dfrac{1}{u}(6x + 10yy')$

$\displaystyle y' - 10yy' = \dfrac{1}{u}(6x)$

$\displaystyle y'[1 - 10y] = \dfrac{1}{u}(6x)$

$\displaystyle y' = \dfrac{1}{u}(6x)\dfrac{1}{(1 - 10y)}$

$\displaystyle y' = \dfrac{1}{3x^{2} + 5y^{2}}(6x)\dfrac{1}{(1 - 10y)}$

$\displaystyle y' = \dfrac{6x}{(3x^{2} + 5y^{2})(1 - 10y)}$ Think this is right.

No, that was wrong. And I have used up all my tries, so I lost point on that question.

How did you get this?

Note: ln(a + b) does not equal ln(a) + ln(b)! :wink:

Thanks for telling me. I didn't realize that.

I am not sure what else to do for these two. Any hints? :?:

How did you get from the above to what you'd entered as your answer?

The answer was my first try.

x^(-1/2)(7x)'

Ridiculous, I know. XD

 

[Jason76]

Ok, can some math guys with a lot of posts help me? Was my answer for problem # 2 sort of right, or on the right track? :?:

How about the 1st one? Assuming the problem was written as I said (but it wasn't) was I using the correct method :?:

 

[Whateverchan]

Was it correct to move the 10yy' to the left like that, though? I think you were multiplying the entire 1/u with 6x + 10yy', so something is fishy there.

 

[Jason76]

Was it correct to move the 10yy' to the left like that, though? I think you were multiplying the entire 1/u with 6x + 10yy', so something is fishy there.

We have to deal with $\displaystyle \dfrac{d}{dx} \ln(u)$ first before doing other stuff. Perhaps your answer came out wrong, because it was simplified too much, or too little. But the procedure I gave was right. Can anyone else back me up on this?

As far moving the $\displaystyle 10yy'$ to the left, that was necessary. Because, in the end, there should be only one $\displaystyle y'$ in the whole equation, and it's supposed to be isolated on the left.

 

[stapel]

I apologize for some of the replies you've received. The other poster isn't familiar much with algebra, so his calculus is usually incorrect.

---

\(\displaystyle \mbox{4. Use logarithmic differentiation to find the derivative of }\, y\, =\, \sqrt{x}^{7x}\)

Start by taking the log of both sides. Then differentiate.

. . . . .$\displaystyle \ln(y)\, =\, \ln\left(\sqrt{x}^{7x}\right)\, =\, \ln\left(x^{\frac{7x}{2}}\right)\, =\, \frac{7x}{2}\, \ln(x)$

. . . . .$\displaystyle \frac{1}{y}\, \frac{dy}{dx}\, =\, \frac{7}{2}\, \ln(x)\, +\, \frac{7x}{2}\, \frac{1}{x}\, =\, \frac{7}{2}\left(\ln(x)\, +\, x\right)$

. . . . .$\displaystyle \frac{dy}{dx}\, =\, \frac{7y}{2}\left(\ln(x)\, +\, x\right)$

Then plug the original expression in for "y" on the right-hand side.

---

\(\displaystyle \mbox{5. Find }\, y'\, \mbox{ if }\, y\, =\, \ln\left(3x^2\, +\, 5y^2\right)\)

Differentiate the log, and implicitly differentiate its argument.

. . . . .$\displaystyle \frac{dy}{dx}\, =\, \left(\frac{1}{3x^2\, +\, 5y^2}\right)\left(6x\, +\, 10y\frac{dy}{dx}\right)$

. . . . .$\displaystyle \frac{dy}{dx}\, =\, \frac{6x}{3x^2\, +\, 5y^2}\, +\, \frac{10y}{3x^2\, +\, 5y^2}\, \frac{dy}{dx}$

. . . . .$\displaystyle \frac{dy}{dx}\, -\, \frac{10y}{3x^2\, +\, 5y^2}\, \frac{dy}{dx}\, =\, \frac{6x}{3x^2\, +\, 5y^2}$

. . . . .$\displaystyle \frac{dy}{dx}\left(1\, -\, \frac{10y}{3x^2\, +\, 5y^2}\right)\, =\, \frac{6x}{3x^2\, +\, 5y^2}$

. . . . .$\displaystyle \frac{dy}{dx}\left(\frac{3x^2\, +\, 5y^2\, -\, 10y}{3x^2\, +\, 5y^2}\right)\, =\, \frac{6x}{3x^2\, +\, 5y^2}$

Divide through and simplify for the derivative.

 

[Whateverchan]

I apologize for some of the replies you've received. The other poster isn't familiar much with algebra, so his calculus is usually incorrect.

---

\(\displaystyle \mbox{4. Use logarithmic differentiation to find the derivative of }\, y\, =\, \sqrt{{x}^{7x}}\)

Start by taking the log of both sides. Then differentiate.

. . . . .$\displaystyle \ln(y)\, =\, \ln\left(\sqrt{{x}^{7x}}\right)\, =\, \ln\left(x^{\frac{7x}{2}}\right)\, =\, \frac{7x}{2}\, \ln(x)$

. . . . .$\displaystyle \frac{1}{y}\, \frac{dy}{dx}\, =\, \frac{7}{2}\, \ln(x)\, +\, \frac{7x}{2}\, \frac{1}{x}\, =\, \frac{7}{2}\left(\ln(x)\, +\, x\right)$

. . . . .$\displaystyle \frac{dy}{dx}\, =\, \frac{7y}{2}\left(\ln(x)\, +\, x\right)$

Then plug the original expression in for "y" on the right-hand side.

---

\(\displaystyle \mbox{5. Find }\, y'\, \mbox{ if }\, y\, =\, \ln\left(3x^2\, +\, 5y^2\right)\)

Differentiate the log, and implicitly differentiate its argument.

. . . . .$\displaystyle \frac{dy}{dx}\, =\, \left(\frac{1}{3x^2\, +\, 5y^2}\right)\left(6x\, +\, 10y\frac{dy}{dx}\right)$

. . . . .$\displaystyle \frac{dy}{dx}\, =\, \frac{6x}{3x^2\, +\, 5y^2}\, +\, \frac{10y}{3x^2\, +\, 5y^2}\, \frac{dy}{dx}$

. . . . .$\displaystyle \frac{dy}{dx}\, -\, \frac{10y}{3x^2\, +\, 5y^2}\, \frac{dy}{dx}\, =\, \frac{6x}{3x^2\, +\, 5y^2}$

. . . . .$\displaystyle \frac{dy}{dx}\left(1\, -\, \frac{10y}{3x^2\, +\, 5y^2}\right)\, =\, \frac{6x}{3x^2\, +\, 5y^2}$

. . . . .$\displaystyle \frac{dy}{dx}\left(\frac{3x^2\, +\, 5y^2\, -\, 10y}{3x^2\, +\, 5y^2}\right)\, =\, \frac{6x}{3x^2\, +\, 5y^2}$

Divide through and simplify for the derivative.

Um, actually, the first one is also wrong because the 7x is outside the square root, not together with x. I used up my last try for that, too. XD

The second one, I am not sure as I can't check, but it looks correct. I appreciate you two for trying to help, though. I couldn't spot the mistake with the last one, so I can't blame him.

 

[pka]

Um, actually, the first one is also wrong because the 7x is outside the square root, not together with x.

That makes no difference.

\(\displaystyle \sqrt{x^{7x}}=x^\left({\dfrac{7x}{2}}\right)=\left(\sqrt x\right)^{7x}\)

 

[stapel]

Um, actually, the first one is also wrong because the 7x is outside the square root, not together with x.

Yeah. Somebody edited my post. I've restored it to its original form.

 

[Deleted member 4993]

Yeah. Somebody edited my post. I've restored it to its original form.

My mistake - that was me -body. There was some comment 7x being inside the √ or outside it - and I was going to explain what pka said - but I got distracted and did not complete my statement.

Since those are equivalent - I thought it "looked better" to have "7x" inside √ sign.

 

[Whateverchan]

My mistake - that was me -body. There was some comment 7x being inside the √ or outside it - and I was going to explain what pka said - but I got distracted and did not complete my statement.

Since those are equivalent - I thought it "looked better" to have "7x" inside √ sign.

So the answer would be the same? Why would it say the answer was wrong, then?

 

[stapel]

So the answer would be the same?

Which answer would be the same as what?

Why would it say the answer was wrong, then?

What is the "it" that "said" "the answer" was wrong? What answer did you supply?

Note: My replies walked you part of the way, and gave instructions for completion. You haven't yet shown your work or answers.

 

[Whateverchan]

Which answer would be the same as what?


What is the "it" that "said" "the answer" was wrong? What answer did you supply?

Note: My replies walked you part of the way, and gave instructions for completion. You haven't yet shown your work or answers.

Would the answer for that question be the same, regardless of whether or not the 7x is outside the radical? The webassign said the answer was wrong.
My answer was ((7sqrt x)^7x / 2) + (ln (x)+x). Yes, I have read what you wrote carefully. I did show my attempt at solving the problems, as poorly as they were.

 

[Deleted member 4993]

Would the answer for that question be the same, regardless of whether or not the 7x is outside the radical? The webassign said the answer was wrong.
My answer was ((7sqrt x)^7x / 2) + (ln (x)+x). Yes, I have read what you wrote carefully. I did show my attempt at solving the problems, as poorly as they were.

That is incorrect - does not match what Stapel suggested.

Work out the last step carefully.

You wrote

$\displaystyle (7\sqrt{x})^7*x/2 + [ln(x) + x]$ ................ and that is incorrect. Please follow PEMDAS (Hierchy of operation) while reporting your answer

Please show us the last step to answer and finally the answer.

 

[Whateverchan]

That is incorrect - does not match what Stapel suggested.

Work out the last step carefully.

You wrote

$\displaystyle (7\sqrt{x})^7*x/2 + [ln(x) + x]$ ................ and that is incorrect. Please follow PEMDAS (Hierchy of operation) while reporting your answer

Please show us the last step to answer and finally the answer.

No, this is what I put. Sorry, I stayed up all night doing other homework so I was sorta half as sleep when I typed.