# Math Help - implicit diff. with logs

1. ## implicit diff. with logs

4x^3 + ln y^2 + 2y = 2x

I don't know what those last 3 terms become after differentiating. The first is 12x^2, but what do those 3 become?

2. Originally Posted by Archduke01
4x^3 + ln y^2 + 2y = 2x

I don't know what those last 3 terms become after differentiating. The first is 12x^2, but what do those 3 become?
It is implied that y is a function of x, Therefore, we treat y as composite (composed with x).

$\ln{y^2}=2\ln{y}$

$\frac{d}{dx}[2\ln{y}]=2\frac{1}{y}\cdot\frac{dy}{dx}$

Do you see my point?

3. Originally Posted by VonNemo19
It is implied that y is a function of x, Therefore, we treat y as composite (composed with x).

$\ln{y^2}=2\ln{y}$

$\frac{d}{dx}[2\ln{y}]=2\frac{1}{y}\cdot\frac{dy}{dx}$

Do you see my point?
Just to verify that I got your explanation, would that mean the derivatives are;

$12x^2 + 2 y'/y + 2yy' = 2$ ?

I'm messing up somewhere in the calculations. I can't seem to get y(1 - 6x^2) / 1 + y which is the given answer.

4. Originally Posted by Archduke01
Just to verify that I got your explanation, would that mean the derivatives are;

$12x^2 + 2 y'/y + 2yy' = 2$ ?

I'm messing up somewhere in the calculations. I can't seem to get y(1 - 6x^2) / 1 + y which is the given answer.
Everythin looks good except the third term on the left hand side. Should be...

$2y'$

not

$2yy'$

5. 12x^2 + 2 y'/y + 2y' = 2
2 y'/y + 2y' = 2 - 12x^2
2y' (1/y + 1) = 2 - 12x^2
y' (1/y + 1)= 1 - 6x^2
y' = (1 - 6x^2) / (y + 1)

Where did I go wrong? The numerator's supposed to have a y in it, but my answer doesn't yield such a thing.

6. Originally Posted by Archduke01
12x^2 + 2 y'/y + 2y' = 2
2 y'/y + 2y' = 2 - 12x^2
2y' (1/y + 1) = 2 - 12x^2
y' (1/y + 1)= 1 - 6x^2

y' = (1 - 6x^2) / (1/y + 1)

multiply numerator and denominator by y ...

y' = y(1 - 6x^2)/(1 + y)

Where did I go wrong? The numerator's supposed to have a y in it, but my answer doesn't yield such a thing.
...

7. Originally Posted by Archduke01
12x^2 + 2 y'/y + 2y' = 2
2 y'/y + 2y' = 2 - 12x^2
2y' (1/y + 1) = 2 - 12x^2
y' (1/y + 1)= 1 - 6x^2
y' = (1 - 6x^2) / (y + 1)

Where did I go wrong? The numerator's supposed to have a y in it, but my answer doesn't yield such a thing.
$\frac{d}{dx}[4x^3+\ln{y^2}+2y]=\frac{d}{dx}[2x]$

$12x^2+\frac{2}{y}y'+2y'=2$

$6x^2+\frac{1}{y}y'+y'=1$

$\frac{1}{y}y'+y'=1-6x^2$

$y'\left(\frac{1}{y}+1\right)=1-6x^2$

$y'=\frac{1-6x^2}{\frac{1}{y}+1}$

$y'=\frac{y-6x^2y}{1+y}$

8. Originally Posted by Archduke01
4x^3 + ln y^2 + 2y = 2x

I don't know what those last 3 terms become after differentiating. The first is 12x^2, but what do those 3 become?
$4x^3 + \ln{y^2} + 2y = 2x$.

$\frac{d}{dx}(4x^3 + \ln{y^2} + 2y) = \frac{d}{dx}(2x)$

$12x^2 + \frac{d}{dx}(\ln{y^2} + 2y) = 2$

$12x^2 + \frac{d}{dy}(\ln{y^2} + 2y)\,\frac{dy}{dx} = 2$

$12x^2 + \left(\frac{2y}{y^2} + 2\right)\,\frac{dy}{dx} = 2$

$\left(\frac{2}{y} + 2\right)\,\frac{dy}{dx} = 2 - 12x^2$

$\frac{dy}{dx} = \frac{2 - 12x^2}{\frac{2}{y} + 2}$

$\frac{dy}{dx} = \frac{1 - 6x^2}{\frac{1}{y} + 1}$.

You could clean this up more if you wanted.

9. Thanks brosephs, I get it now. Damn, that 2 layered denominator was confusing.

10. Originally Posted by Archduke01
But if we're seeing y as a function, shouldn't we have to apply the chain rule?
Sure! That's what we did!

$\frac{d}{dx}2y=(1)2y^{1-1}y'=2(1)y'=2y'$

11. Originally Posted by VonNemo19
Sure! That's what we did!

$\frac{d}{dx}2y=2y^{1-1}y'=2(1)y'=2y'$
XD yeah I realized that just after I posted. My badness.