Understanding Implicit Differentiation: Solving for Second Order Derivatives

[Duncan1382]

Find [tex]d^2/dx^2(3y^2+8y=3x)[/tex]

I managed to get [tex]dy/dx = 3 / (6y + 8)[/tex] but I have no clue where to go from here.

According to WolfRamAlpha, the answer is [tex]-27/(4(16 + 9x)(4 + 3y))[/tex], but since dy/dx doesn't have any x value in it, I don't see how the derivative of it would.

I've played around with it for a long time, and I just can't get it. Help please?

 

[semc]

Do you need to express in terms of x and y? I don't see what's wrong if you express it in y which is what i would do

 

[Duncan1382]

Do you need to express in terms of x and y? I don't see what's wrong if you express it in y which is what i would do

You can't express y in terms of x. It's not a function. You can express it in terms of x and y, but that just makes the math longer.

 

[TheoMcCloskey]

Duncan - WolfRamAlpha mad a substitution in the denominator using the original expression.

Firstly - did you compute the second derivative correctly? You should have yielded:

[tex]
y' =\frac{3}{8+6 \cdot y}
[/tex]
and
[tex]
y'' = -\frac{6 \cdot (y')^2}{8+6 \cdot y}
[/tex]

Now, using the first into the second

[tex]
y'' = -\left (\frac{3}{8+6 \cdot y} \right)^2 \left (\frac{6}{8+6 \cdot y} \right ) = \frac{-54}{8 \cdot (4+3 \cdot y)^3}
[/tex]

Now, using the fact that [itex]3y^2+8y=3x[/itex], you need to show yourself that [itex](4+3y)^2 = 9x +16[/itex]. Use this in above to yield result.

 

[Duncan1382]

Oh. Now I get it.

Thank you.