Using chain rule to derive 2nd derivative

[mrcleanhands]

Homework Statement

Use \frac{\partial z}{\partial r}=\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial y}
and \frac{\partial z}{\partial\theta}=-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y} to show that

\frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}+\frac{1}{r} \frac{\partial z}{\partial r}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial ^ {2}z}{\partial x \partial y}+\cos^{2}\theta\frac{\partial ^ {2}z}{\partial y^{2}}

Homework Equations

The Attempt at a Solution

<br /> \frac{\partial z}{\partial\theta}\frac{\partial z}{\partial\theta}=\frac{\partial z}{\partial\theta}(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})<br />

<br /> \frac{\partial^{2}z}{\partial\theta^{2}}=(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})

<br /> \frac{\partial^{2}z}{\partial\theta^{2}}=r^{2}\sin^{2}\theta\frac{ \partial ^ {2}z}{\partial x^{2}}-2r^{2}\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+r^{2}\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}

<br /> \frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}

but \frac{1}{r}\frac{\partial z}{\partial r}=\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y}

and if I add that I get:
\frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}+\frac{1}{r} \frac{ \partial z}{\partial r}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}+\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y} so somehow \frac{1}{r}\frac{\partial z}{\partial r}=\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y} is supposed to be 0?

 

[HallsofIvy]

Homework Statement

Use \frac{\partial z}{\partial r}=\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial y}
and \frac{\partial z}{\partial\theta}=-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y} to show that

\frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}+\frac{1}{r} \frac{\partial z}{\partial r}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial ^ {2}z}{\partial x \partial y}+\cos^{2}\theta\frac{\partial ^ {2}z}{\partial y^{2}}

Homework Equations

The Attempt at a Solution

<br /> \frac{\partial z}{\partial\theta}\frac{\partial z}{\partial\theta}=\frac{\partial z}{\partial\theta}(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})<br />

<br /> \frac{\partial^{2}z}{\partial\theta^{2}}=(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})

What? When I saw "\frac{dz}{d\theta}\frac{dz}{d\theta}" above I started to write "No, that's the wrong notation- that means the product" but now it appears that you really are just multiplying the first derivative with itself. You do understand that this is NOT what "\frac{d^2z}{d\theta^2}&quot; means, don&#039;t you?<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> &lt;br /&gt; \frac{\partial^{2}z}{\partial\theta^{2}}=r^{2}\sin^{2}\theta\frac{ \partial ^ {2}z}{\partial x^{2}}-2r^{2}\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+r^{2}\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}<br /> <br /> &lt;br /&gt; \frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}<br /> <br /> but \frac{1}{r}\frac{\partial z}{\partial r}=\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y}<br /> <br /> and if I add that I get:<br /> \frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}+\frac{1}{r} \frac{ \partial z}{\partial r}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}+\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y} so somehow \frac{1}{r}\frac{\partial z}{\partial r}=\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y} is supposed to be 0? </div> </div> </blockquote>