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So I have the equation

$$\tan^2 \theta = \frac{x^2+y^2}{z^2}$$

I got the first derivative of $\theta$ with respect to $x$ by doing this

\begin{align} 2\sec^2 \theta \tan \theta d \theta &= \frac{2x}{z^2} dx \\ \frac{d \theta}{dx} &= \frac{x \cos^3 \theta}{z^2 \sin \theta} \end{align}

But I'm having difficulty solving for $\frac{d^2\theta}{dx^2}$. Any help?

Matthew Cassell
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mEXsACHINE
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  • $\displaystyle{\frac{d^2\theta}{dx^2}=\frac{d}{dx}\frac{d\theta}{dx}}$ – Evan Tseng Dec 05 '22 at 11:07
  • Differentiate again and put the value of $d\theta /dx$ and then express everything in $tan\theta$. Replace the value of $tan\theta$ from the question to get the answer. – Vikash Kumar Dec 05 '22 at 11:07

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