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Question Number 212464 by liuxinnan last updated on 14/Oct/24

Commented by liuxinnan last updated on 14/Oct/24

$$isitright$$

Answered by MrGaster last updated on 14/Oct/24

$$\mathrm{The}\mathrm{above}\mathrm{process}\mathrm{is}.$$$$\mathrm{correct}\mathrm{Let}\mathrm{me}\mathrm{explain}\mathrm{ite}$$$$\mathrm{her}.$$$$\mathrm{Firstly}\mathrm{given}\mathrm{the}\mathrm{functionalr}$$$$\mathrm{elationship}y=f\left(x\right)\mathrm{and}\mathrm{its}\mathrm{inverse}$$$$\mathrm{function}x=\phi \left(y\right).$$$$\mathrm{According}\mathrm{to}\mathrm{the}$$$$\mathrm{derivativeformula}\mathrm{fore}$$$$\mathrm{invers}\mathrm{functions}\mathrm{we}\mathrm{knowa}$$$$\mathrm{tht}\mathrm{if}y=f\left(x\right),\mathrm{then}{f}^{\prime }\left(x\right)=\frac{dy}{dx}\mathrm{Similarly}\mathrm{for}\mathrm{the}\mathrm{inversec}$$$$\mathrm{funtion}x=\phi \left(y\right),\mathrm{we}\mathrm{have}$$$${\phi }^{\prime }\left(y\right)=\frac{dx}{dy}$$$$S\mathrm{ince}x\mathrm{and}y\mathrm{are}\mathrm{in}\mathrm{a}\mathrm{reciprocali}$$$$\mathrm{relationshp}\mathrm{as}\mathrm{inversei}$$$$\mathrm{functons},\frac{dy}{dx}\mathrm{and}\frac{dx}{dy}\mathrm{are}\mathrm{reciprocals}\mathrm{of}\mathrm{each}$$$$\mathrm{otherwhich}\mathrm{means}:$$$${\phi }^{\prime }\left(x\right)=\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}=\frac{1}{f^{\prime }\left(x\right)}$$$$\mathrm{Next},\mathrm{since}x=\phi \left(y\right),\mathrm{we}\mathrm{can}\mathrm{substitute}x\mathrm{with}\phi \left(y\right),\mathrm{thus}\mathrm{obtaining}:$$$${\phi }^{\prime }\left(y\right)=\frac{1}{f^{\prime }\left[\phi \left(y\right)\right]}$$

Commented by liuxinnan last updated on 15/Oct/24

$$thanks$$

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