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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?

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    $\begingroup$ Please try not to contain images in your post. Some users may not able to see them due to various reasons. $\endgroup$ – xbh Aug 28 '18 at 12:23
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    $\begingroup$ Note that $y = y(x)$ is a function of $x$, so… $\endgroup$ – xbh Aug 28 '18 at 12:24
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    $\begingroup$ It seems there is amistake for $(x,y)=(2,1)$, I thnk it should be $(x,y)=(1,2)$ to satisfy the equation. I fix that $\endgroup$ – gimusi Aug 28 '18 at 12:34
  • $\begingroup$ Could equally be $(4,1)$! $\endgroup$ – preferred_anon Aug 28 '18 at 14:38
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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 $(x,y)=(1,2)$

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

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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!

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  • $\begingroup$ It seems there is amistake for $(x,y)=(2,1)$, I thnk it should be $(x,y)=(1,2)$ to satisfy the equation. $\endgroup$ – gimusi Aug 28 '18 at 12:34
  • $\begingroup$ @gimusi Yes I saw that right now. He might made a mistake when typing the question. $\endgroup$ – Anastassis Kapetanakis Aug 28 '18 at 12:36
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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

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  • $\begingroup$ When you implicitly differentiate $$y^4+5x=21$$ you get $$4y^3\frac{dy}{dx}+5=0$$ I then differentiated the $4y^3\frac{dy}{dx}$ using the product rule, and the $+5$ disappears. $\endgroup$ – Rhys Hughes Aug 29 '18 at 13:37
  • $\begingroup$ You can differentiate $\frac{-5}{4y^3}$ with $p=-5\to p'=0$ and $q=4y^3\to q'=12y^2$. Then the quotient rule is: $$\frac{d}{dx}(\frac pq)=\frac{p'q-q'p}{q^2}$$ which leads to $$\frac{0-60y^2}{16y^6}=\frac{-15}{4y^4}$$ $\endgroup$ – Rhys Hughes Aug 29 '18 at 13:43

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