Question Number 212464 by liuxinnan last updated on 14/Oct/24
Commented by liuxinnan last updated on 14/Oct/24
$${is}\:{it}\:{right} \\ $$
Answered by MrGaster last updated on 14/Oct/24
![The above process is. correct Let me explain ite her. Firstly given the functionalr elationship y=f(x)and its inverse function x=ϕ(y). According to the derivativeformula fore invers functions we knowa tht if y=f(x),then f′(x)=(dy/dx)Similarly for the inversec funtion x=ϕ(y),we have ϕ′(y)=(dx/dy) Since x and y are in a reciprocali relationshp as inversei functons ,(dy/dx)and(dx/dy)are reciprocals of each otherwhich means: ϕ′(x)=(dx/dy)=(1/(dy/dx))=(1/(f′(x))) Next,since x=ϕ(y),we can substitute x with ϕ(y),thus obtaining: ϕ′(y)=(1/(f′[ϕ(y)]))](https://www.tinkutara.com/question/Q212466.png)
$$\mathrm{The}\:\mathrm{above}\:\mathrm{process}\:\mathrm{is}. \\ $$$$\mathrm{correct}\:\mathrm{Let}\:\mathrm{me}\:\mathrm{explain}\:\mathrm{ite} \\ $$$$\mathrm{her}. \\ $$$$\mathrm{Firstly}\:\mathrm{given}\:\mathrm{the}\:\mathrm{functionalr} \\ $$$$\mathrm{elationship}\:{y}={f}\left({x}\right)\mathrm{and}\:\mathrm{its}\:\mathrm{inverse} \\ $$$$\mathrm{function}\:{x}=\varphi\left({y}\right). \\ $$$$\mathrm{According}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{derivativeformula}\:\mathrm{fore} \\ $$$$\mathrm{invers}\:\mathrm{functions}\:\mathrm{we}\:\mathrm{knowa} \\ $$$$\mathrm{tht}\:\mathrm{if}\:{y}={f}\left({x}\right),\mathrm{then}\:{f}'\left({x}\right)=\frac{{dy}}{{dx}}\mathrm{Similarly}\:\mathrm{for}\:\mathrm{the}\:\mathrm{inversec} \\ $$$$\mathrm{funtion}\:{x}=\varphi\left({y}\right),\mathrm{we}\:\mathrm{have}\: \\ $$$$\varphi'\left({y}\right)=\frac{{dx}}{{dy}} \\ $$$${S}\mathrm{ince}\:{x}\:\mathrm{and}\:{y}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a}\:\mathrm{reciprocali} \\ $$$$\mathrm{relationshp}\:\mathrm{as}\:\mathrm{inversei} \\ $$$$\mathrm{functons}\:,\frac{{dy}}{{dx}}\mathrm{and}\frac{{dx}}{{dy}}\mathrm{are}\:\mathrm{reciprocals}\:\mathrm{of}\:\mathrm{each}\: \\ $$$$\mathrm{otherwhich}\:\mathrm{means}: \\ $$$$\varphi'\left({x}\right)=\frac{{dx}}{{dy}}=\frac{\mathrm{1}}{\frac{{dy}}{{dx}}}=\frac{\mathrm{1}}{{f}'\left({x}\right)} \\ $$$$\mathrm{Next},\mathrm{since}\:\:{x}=\varphi\left({y}\right),\mathrm{we}\:\mathrm{can}\:\mathrm{substitute}\:{x}\:\mathrm{with}\:\varphi\left({y}\right),\mathrm{thus}\:\mathrm{obtaining}: \\ $$$$\varphi'\left({y}\right)=\frac{\mathrm{1}}{{f}'\left[\varphi\left({y}\right)\right]} \\ $$