Show that the Second Derivative of the equation can be written as:

[RyanGall93]

Hi, need a bit of help working through the steps of some second order differentiation given to me as a part of a tutorial for an upcoming exam. Thanks

 

[tkhunny]

First, equations don't have derivatives.
Second, have you found dy/dx?

 

[RyanGall93]

Yeah, I've got dy/dx as 2X + 2Y - 3 = 0

 

[Dr.Peterson]

That isn't dy/dx; you appear to have differentiated the equation implicitly while forgetting the chain rule.

Use the chain rule, and you can then solve for dy/dx.

 

[tkhunny]

Yeah, I've got dy/dx as 2X + 2Y - 3 = 0

One normally assumes y = f(x). It should be in the instructions or the section header for the problem.

 

[Steven G]

Yeah, I've got dy/dx as 2X + 2Y - 3 = 0

What does that mean? dy/dx is NOT an equation. When you are asked what dy/dx equals your response should be in the form dy/dx = ....

 

[pka]

Find the second derivative\(\displaystyle \frac{d^2y}{dx^2}\) of \(\displaystyle x^2+y^2-3y=5\).
Well \(\displaystyle \begin{align*}2x+2y\,y'&-3y'=0 \\y'&=\frac{-2x}{(2y-3)} \end{align*}\)
Then we get:
\(\displaystyle \begin{align*}y''&=\frac{-2(2x-3)+(2x)(2y')}{(2y-3)^2} \\&=\frac{-2(2x-3)+4xy'}{(2y-3)^2}\\&=\frac{-2(2x-3)+4x\left(\frac{-2x}{(2y-3)}\right)}{(2y-3)^2}\\&=\frac{-2(2y-3)^2-8x^2}{(2y-3)^3}\\&=\frac{-2(4y^2-12y+9)-8x^2}{(2y-3)^3}\\&=\frac{-8(x^2+y^2-3y)-18}{(2y-3)^3}\\&=\frac{-8(5)-18}{(2y-3)^3} \end{align*}\\\text{QED}\)

 

[JeffM]

There is a minor but distracting typo in pka's post.

In the line after "then we get," he meant (2y - 3) rather than (2x - 3) in the first term of the numerator