How to take second derivative implicitly

Question

Let $$y^4 + 5x = 21.$$ What is the value of $d^2y/dx^2$ at the point $(2, 1)$?

I’m stuck at trying to work out the second implicit derivative of this function. As far as I can work out, the first derivative implicitly is

$$\dfrac {-5}{4y^{3}}$$

How do you take the second derivative implicitly with respect to $x$ when $x$ has vanished? There would be nowhere to plug in $x=2$. What am I missing here?

Answer 1

We have

$$y^4+5x=21 \implies 4y^3dy+5dx=0 \quad y'=\frac{-5}{4y^3}$$

therefore by chain rule

$$y''=\frac{dy'}{dy}\frac{dy}{dx}=\frac{15}{4y^4}y'$$

Now it seems there is a mistake for $(x,y)=(2,1)$, it should be $(x,y)=(1,2)$ in order to satisfy the equation, then and at $(x,y)=(1,2)$

• $y'(1)=\frac{-5}{4\cdot 2^3}=-\frac5{32}$
• $y''(1)=\frac{15}{4\cdot 2^4}\cdot \left(-\frac5{32}\right)=-\frac{75}{2048}$

Answer 2

You just plugin in the first equation the value $x=1$ and solve for $y$. Then in the derivative you found you put in $y$ the value you got at the first step and you can find the desired value.

Furthermore, in order to find the second derivative you must take into account the chain rule. Then you will have: $$y''=\frac{15}{4y^4}y'$$

Then you plugin the values of $y'$ and $y$ from before and you are done!

I wish I helped!

Answer 3

There's a typo in your problem statement. The point $(2,1)$ is not on the curve. They probably meant $(1,2)$.

I would not have solved for the first derivative. Just differentiate the equation again, remembering that both $y$ and $y'$ are functions of $x$, and so the chain rule applies:

$$y^4+5x=21$$

$$4y^3y'+5=0$$

$$4y^3y'' + 12y^2y'y' = 0.$$

Plug in $y=2$ in the second line to get $4\cdot 8 y'+5=0$ and solve to get $y' = -5/32.$

Then plut $y=2$ and $y'=-5/32$ in the third line and solve for $y''$. I get $-75/2048.$ (But I've had only one coffee this morning.)

Answer 4

You got: $$\frac{dy}{dx}=\frac{-5}{4y^3}$$ This is correct. Now notice the product rule: $$\frac{d}{dx}(pq)=p'q+pq'$$ Here: $$p=4y^3\to p'=12y^2\frac{dy}{dx}$$ $$q=\frac{dy}{dx}\to q'=\frac{d^2y}{dx^2}$$ Thus via implicit differentiation we have:

$$4y^3\frac{d^2y}{dx^2}+12y^2(\frac{dy}{dx})^2=0$$ $$4y^3\frac{d^2y}{dx^2}-\frac{15}{y}=0$$ and go from there