# Math Help - Find Second Derivative Implicitly

1. ## Find Second Derivative Implicitly

I am learning calculus 1 on my own. Currently, I am studying implicit differentiation. I am also reviewing a few topics from precalculus. I have several questions to post.

For the first question, I need to find the second derivative. The answer is (3x/4y). I found y' to be 3x^2/(2y). I know the quotient rule is needed to find the second derivative. Look at the picture below. How do I find the second derivative via quotient rule?

2. ## Re: Find Second Derivative Implicitly

It would be easier for you after you have gotten to \displaystyle \begin{align*} y' = \frac{3x^2}{2y} \end{align*} to write it as \displaystyle \begin{align*} y' = \frac{3x^2}{2x^{\frac{3}{2}}} = \frac{3}{2}x^{\frac{1}{2}} \end{align*}. This is very easy to differentiate.

3. ## Re: Find Second Derivative Implicitly

Where did 2y go to?
I do not see it in your simplified fraction.

4. ## Re: Find Second Derivative Implicitly

Hello, nycmath!

$\text{Given: }\:y^2\,=\,x^3\;\;{\color{red}[1]}\quad \text{ Find }\,y''.$

$\text{I found: }\:y' \;=\;\frac{3x^2}{2y}\;\;{\color{red}[2]}$ . Correct!

$\text{I got as far as: }\:y'' \;=\;\frac{2y(6x) - 3x(2y')}{(2y)^2}$ . Good!

Now we do a lot of substituting and simplifying . . .

$\text{We have: }\:y'' \;=\;\frac{12xy - 6x^2y'}{4y^2}$

$\text{Substitute }{\color{red}[2]}:\;y'' \;=\;\frac{12xy - 6x^2\left(\frac{3x^2}{2y}\right)}{4y^2} \;=\;\frac{12xy - \frac{9x^4}{y}}{4y^2}$

$\text{Multiply by }\frac{y}{y}\!:\;y'' \;=\;\frac{12xy^2 - 9x^4}{4y^3} \;=\;\frac{3x(4y^2-3x^3)}{4y^2}$

$\text{Substitute }{\color{red}[1]}\!:\;y'' \;=\;\frac{3x(4y^2-3\!\cdot\!y^2)}{4y^3} \;=\;\frac{3xy^2}{4y^3}$

$\text{Therefore: }\:y'' \;=\;\frac{3x}{4y}$

5. ## Re: Find Second Derivative Implicitly

Thank you, Soroban. A few minutes before your post, I was able to find the answer. It involves lots of substituting and simplification. The process is a bit long but fun.

6. ## Re: Find Second Derivative Implicitly

Originally Posted by nycmath
Where did 2y go to?
I do not see it in your simplified fraction.
Your original function was \displaystyle \begin{align*} y^2 = x^3 \implies y = x^{\frac{3}{2}} \end{align*}, so I just replaced the \displaystyle \begin{align*} y \end{align*} with \displaystyle \begin{align*} x^{\frac{3}{2}} \end{align*}.

Work smart, not hard...

7. ## Re: Find Second Derivative Implicitly

Thank you, Prove it. Nice trick.