# derivative of function

I have a simple problem from calculus topics. Suppose we have
$$x=at^2,\qquad y=2at$$ and want to find $\dfrac{d^2y}{dx^2}$. There is given sample http://www.mathopolis.com/questions/a.php?id=137&ansno=957 I think that the answer is zero but here it is equal D why?

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You are not taking the derivative of $y$ with respect to $t$, you are taking the derivative of $y$ with respect to $x$.

Now, by the Chain Rule we have that $$\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt},$$ or, equivalently (solving for $\frac{dy}{dx}$), $$\frac{dy}{dx} = \frac{\quad\frac{dy}{dt}\quad}{\frac{dx}{dt}}.$$ Therefore, since $\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a$, and $\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$, then $$\frac{dy}{dx} = \frac{2a}{2at} = t^{-1}.$$

Now you can repeat: from the Chain Rule we have $$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \Biggl(\frac{d}{dx}\left(\frac{dy}{dx}\right)\Biggr)\frac{dx}{dt} = \frac{d^2y}{dx^2}\frac{dx}{dt}.$$ Solving for $\frac{d^2y}{dx^2}$, we get $$\frac{d^2y}{dx^2} = \frac{\quad\frac{d}{dt}\left(\frac{dy}{dx}\right)\quad}{\frac{dx}{dt}};$$ since $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}(t^{-1}) = -t^{-2}$, we have: $$\frac{d^2y}{dx^2} = \frac{-t^{-2}}{2at} = -\frac{1}{2at^3},$$ which is answer D in the given link.

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Observe that $y^2 = 4ax$

Then $2y\frac{dy}{dx} = 4a$

So $\frac{dy}{dx} = \frac{2a}{y}$ [A]

Then $\frac{\mathrm d^2 y}{\mathrm dx^2} = \frac{-2a}{y^2} \frac{dy}{dx} =$ (using [A]) $-\frac{4a^2}{y^3} = -\frac{4a^2}{8a^3t^3} = -\frac{1}{2at^3}$

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A slightly different approach not using the chain rule explicitly:

From $x=at^2$ you have $$t=a^{-1/2}x^{1/2}$$ so with $y=2at$ you have $$y =2 a^{1/2}x^{1/2}$$ so taking the derivative with respect to $x$ $$\frac{\mathrm dy}{\mathrm dx}=a^{1/2}x^{-1/2}$$ and doing it again and reusing $x=at^2$ $$\frac{\mathrm d^2 y}{\mathrm dx^2}=-\frac{1}{2} a^{1/2}x^{-3/2}=-\frac{1}{2a t^3}.$$

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Minor point: you assumed $a > 0$. –  Shai Covo Jun 29 '11 at 10:37
@Shai: True. Note that if $a \lt 0$ then $x \le 0$ so to deal with the point you would get $t=\sqrt{x/a}$, $y=2a\sqrt{x/a}$, $\mathrm dy / \mathrm dx =a\sqrt{1/ax}$ and $\mathrm d^2 y /\mathrm dx^2 =-\frac{1}{2} a\sqrt{1/ax^3} =-\frac{1}{2a t^3}$. –  Henry Jun 29 '11 at 11:08
Thanks for this response. –  Shai Covo Jun 29 '11 at 11:16

Here, you use the fact that

$$\frac{\mathrm dy}{\mathrm dx}=\frac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$$

Since you're doing second derivatives, you need a further differentiation:

$$\frac{\mathrm d^2 y}{\mathrm dx^2}=\left(\frac{\mathrm dx}{\mathrm dt}\right)^{-1}\frac{\mathrm d}{\mathrm dt}\frac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$$

and this is the formula you should be using (replace $x$ and $y$ with the appropriate expressions).

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I'm not sure what else to add to the explanation provided in your link. I think the key to understanding this question is the statement $$\frac{dy}{dx}=\frac{\quad\tfrac{dy}{dt}\quad}{\tfrac{dx}{dt}},$$ which follows from the chain rule: $$\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}.$$ Do you understand why this is true?

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look this i know but let express a as x/t^2 and put into second equation i have got y=2*x/t^2 –  dato datuashvili Jun 29 '11 at 8:01
That's a valid manipulation, but when you compute $$\frac{dy}{dx}=\frac{d}{dx}\left(2at\right)=\frac{d}{dx}\left(2\cdot \frac{x}{t^2} \cdot t\right)=\frac{d}{dx}\left(\frac{2x}{t}\right)$$ you now have to treat $t$ as a function of $x$, not as a constant. –  Zev Chonoles Jun 29 '11 at 8:04