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Differentiating dy/dα w.r.t x what would we get? The textbook i'm using says the answer is (d^2y/dα^2)(dα/dx) but I'm not able to understand it.

Sidd
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  • It's just chain rule: you have $y=f(\alpha)$ and ${dy\over d\alpha}=f'(\alpha)$. Then, assuming $\alpha=g(x)$, use the chain rule ${dq \over dx}={dq \over dz}{dz \over dx}$ to obtain ${d \over dx} {dy\over d\alpha}= {d \over dx}f'(g(x))=f''(g(x))g'(x)= {d^2y\over d\alpha^2}{d\alpha \over dx}$ – N74 May 06 '17 at 10:28

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What they are doing is that they are setting dy/da to another variable and then differentiating that variable using implicit differentiation. For understanding the technique, you should check out this website, https://revisionmaths.com/advanced-level-maths-revision/pure-maths/calculus/implicit-differentiation. If the answer is not helpful, please inform and I will try my best to provide a better answer. Please don't give any down votes.

Soumil
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