If the solve the problem


If $x=4 z^{2}+5, y=6 z^{2}+7 z+3$, find $\frac{d^{2} y}{d x^{2}}$


Formula: -

(i) $\frac{d y}{d x}=y_{1}$ and $\frac{d^{2} y}{d x^{2}}=y_{2}$

(ii) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{\mathrm{n}}=\mathrm{nx}^{\mathrm{n}-1}$

(iii) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d}(\text { wou })}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

(iv) parameteric forms $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Given: -

$x=4 z^{2}+5, y=6 z^{2}+72+3$

Differentiating both w.r.t z

$\frac{\mathrm{dx}}{\mathrm{dz}}=8 \mathrm{z}+0$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dz}}=\frac{12 \mathrm{z}+7}{8 \mathrm{z}}$

and $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dz}}=12 \mathrm{z}+7$

differentiating w.r.t z

$\frac{d\left(\frac{d y}{d x}\right)}{d z}=0+\frac{7}{8}\left(\frac{-1}{z^{2}}\right)$


$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{-7}{8 \mathrm{z}^{2} \times 8 \mathrm{z}}=\frac{-7}{64 \mathrm{z}^{3}}$

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