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If $x=f(\theta)$ and $y=g(\theta)$, then $\frac{d^2 y}{d x^2}=$
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Verified Answer
The correct answer is:
$\frac{f^{\prime}(\theta) g^{\prime \prime}(\theta)-g^{\prime}(\theta) f^{\prime \prime}(\theta)}{\left(f^{\prime}(\theta)\right)^3}$
Given,
$x=f(\theta)$
and $y=g(\theta)$
$\frac{d x}{d \theta}=f^{\prime}(\theta), \frac{d y}{d \theta}=g^{\prime}(\theta)$
$\Rightarrow \quad \frac{d y}{d x}=\frac{g^{\prime}(\theta)}{f^{\prime}(\theta)}$
$\begin{aligned} & \Rightarrow \frac{d^2 y}{d x^2}=\left(\frac{g^{\prime \prime}(\theta) f^{\prime}(\theta)-f^{\prime \prime}(\theta) g^{\prime}(\theta)}{\left(f^{\prime}(\theta)^2\right)}\right) \frac{d \theta}{d x} \\ & \Rightarrow \frac{d^2 y}{d x^2}=\left(\frac{g^{\prime \prime}(\theta) f^{\prime}(\theta)-f^{\prime \prime}(\theta) g^{\prime}(\theta)}{\left(f^{\prime}(\theta)\right)^3}\right)\end{aligned}$
$x=f(\theta)$
and $y=g(\theta)$
$\frac{d x}{d \theta}=f^{\prime}(\theta), \frac{d y}{d \theta}=g^{\prime}(\theta)$
$\Rightarrow \quad \frac{d y}{d x}=\frac{g^{\prime}(\theta)}{f^{\prime}(\theta)}$
$\begin{aligned} & \Rightarrow \frac{d^2 y}{d x^2}=\left(\frac{g^{\prime \prime}(\theta) f^{\prime}(\theta)-f^{\prime \prime}(\theta) g^{\prime}(\theta)}{\left(f^{\prime}(\theta)^2\right)}\right) \frac{d \theta}{d x} \\ & \Rightarrow \frac{d^2 y}{d x^2}=\left(\frac{g^{\prime \prime}(\theta) f^{\prime}(\theta)-f^{\prime \prime}(\theta) g^{\prime}(\theta)}{\left(f^{\prime}(\theta)\right)^3}\right)\end{aligned}$
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