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When we're given some function, say y = $\sin^2\theta$ and x = $\cos^2\theta$, to find $\frac{dy}{dx}$, we use the following way:

$\frac{dy}{d\theta}$ = $2\sin\theta\cos\theta$ $\frac{dx}{d\theta}$ = -$2\sin\theta\cos\theta$

giving us:

$\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{dy}{dx} = -1$

Then, to find $\frac{d^2y}{dx^2}$ we use chain rule giving us 0.

Instead of doing that, can we differentiate y and x with respect to theta twice and divide those to get $\frac{d^2y}{dx^2}$? Why or why not?

My point of confusion is that if we can't, then why is this wrong:

We assume $\frac{dy}{d\theta} = k$ and $\frac{dx}{d\theta} = l$ as new functions and then differentiate those again and divide the results, plug the aforementioned in and arrive at $\frac{d^2y}{dx^2}$.

Utkarsh Jetly
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    No, you can't differentiate both twice wrt $\theta$ and divide these to get $d^2y/dx^2$ - see here – A. Goodier Aug 15 '22 at 20:38
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    Also, for your example it is probably easier to just note that $y=1-x$, which enables you to work out the derivatives more quickly. – A. Goodier Aug 15 '22 at 20:38
  • To understand how to deal with second differentials algebraically, see "Extending the Algebraic Manipulability of Differentials". https://www.academia.edu/37619039/Extending_the_Algebraic_Manipulability_of_Differentials – johnnyb Aug 16 '22 at 02:05

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Let $(\Delta_{h,x})( f,x)=f(x+h)-f(h).$ Then $(\Delta_{h,x})^2(f,x)=f(x+2h)-2f(x+h)+f(x).$

If the 2nd derivative of $f$ with respect to $x$ is continuous then $$\frac {(\Delta_{h,x})^2(f,x)}{h^2}=\frac {(\Delta_{h,x})^2(f,x)}{[(\Delta_{h,x}(id_x,x)[^2}$$ converges to it as $h\to 0.$ This is why we write $\frac {d^2f}{(dx)^2}.$

But $$\frac {d^2y(\theta)}{(d \theta)^2}\;/\; \frac {d^2x(\theta)}{(d\theta)^2}=$$ $$=\lim_{h\to 0}\frac {(\Delta_{h,\theta})^2(y,\theta)}{(\Delta_{h,\theta})^2(x,\theta)}$$ while $$\frac {d^2y}{dx^2}=\lim_{h\to 0}\frac {(\Delta_{h,\theta})^2(y,\theta)}{[(\Delta_{h,\theta}(x,\theta)]^2}$$ and these are usually two different things. E.g. if $x=\theta^2$ and $y=\theta^4$ then $y=x^2$ so $\frac {d^2y}{dx^2}=2$ but $\frac {d^2y}{d\theta^2}\;/\;\frac {d^2x}{d\theta^2}=12\theta^2/2=6\theta^2=6x.$

DanielWainfleet
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  • If $y= sin^2(\theta)$ and $x= cos^2(\theta)$ then $x+ y= cos^2(\theta)+ sin^2(\theta)= 1$ so $y= 1- x$, $\frac{dy}{dx}= -1$ and $\frac{d^2y}{dx^2}= 0$. – George Ivey Aug 15 '22 at 23:01
  • @GeorgeIvey. Yes. The proposer noted that $d^2y/dx^2\ne (d^2y/d\theta^2)/(d^2x/d\theta^2)$ and wanted to know why, especially as a somewhat similar method gives a valid result $dy/dx=(dy/d\theta)/(dx/d\theta)$ for the 1st derivative. – DanielWainfleet Aug 17 '22 at 00:15