Question

x^1009y^1011=(x+y)^2020 dy÷dx=​

Answer 1

(1) Direct derivative:

Solution: $\mathrm{x^{1009}.y^{1011}=(x+y)^{2020}}$ ... ...(1)

Differentiating both sides with respect to $\mathrm{x},$ we get

$\quad \mathrm{\dfrac{d}{dx}(x^{1009}.y^{1011})=\dfrac{d}{dx}[(x+y)^{2020}]}$

$\mathrm{\Rightarrow 1011.x^{1009}.y^{1010}.\dfrac{dy}{dx}+1009.x^{1008}.y^{1011}=2020.(x+y)^{2019}.(1+\dfrac{dy}{dx})}$

$\mathrm{\Rightarrow\dfrac{1011.x^{1009}.y^{1011}}{y}.\dfrac{dy}{dx}+\dfrac{1009.x^{1009}.y^{1011}}{x}=2020.(x+y)^{2019}.(1+\dfrac{dy}{dx})}$

$\mathrm{\Rightarrow\dfrac{1011}{y}.(x+y)^{2020}.\dfrac{dy}{dx}+\dfrac{1009}{x}.(x+y)^{2020}=2020.(x+y)^{2019}.(1+\dfrac{dy}{dx})\:[by\:(1)]}$

$\mathrm{\Rightarrow\dfrac{1011}{y}.(x+y).\dfrac{dy}{dx}+\dfrac{1009}{x}.(x+y)=2020.(1+\dfrac{dy}{dx})}$

$\mathrm{\Rightarrow[\dfrac{1011}{y}.(x+y)-2020].\dfrac{dy}{dx}=2020-\dfrac{1009}{x}.(x+y)}$

$\mathrm{\Rightarrow\dfrac{1011x+1011y-2020y}{y}.\dfrac{dy}{dx}=\dfrac{2020x-1009x-1009y}{x}}$

$\mathrm{\Rightarrow\dfrac{1011x-1009y}{y}.\dfrac{dy}{dx}=\dfrac{2020x-1009y}{x}}$

$\mathrm{\Rightarrow\dfrac{dy}{dx}=\dfrac{y}{x}}$

This is the required derivative.

(2) Logarithmic derivative:

Solution: $\mathrm{x^{1009}.y^{1011}=(x+y)^{2020}}$

Taking $\mathrm{log}$ to both sides, we get

$\quad \mathrm{log(x^{1009}.y^{1011})=log[(x+y)^{2020}]}$

$\mathrm{\Rightarrow log(x^{1009})+log(y^{1011})=log[(x+y)^{2020}]}$

$\mathrm{\Rightarrow 1009.logx+1011.logy=2020.log(x+y)}$

Differentiating both sides with respect to $\mathrm{x},$ we get

$\quad \mathrm{\dfrac{d}{dx}(1009.logx+1011.logy)=\dfrac{d}{dx}[2020.log(x+y)]}$

$\mathrm{\Rightarrow 1009.\dfrac{d}{dx}(logx)+1011.\dfrac{d}{dx}(logy)=2020.\dfrac{d}{dx}[log(x+y)]}$

$\mathrm{\Rightarrow\dfrac{1009}{x}+\dfrac{1011}{y}.\dfrac{dy}{dx}=\dfrac{2020}{x+y}.(1+\dfrac{dy}{dx})}$

$\mathrm{\Rightarrow\dfrac{1011}{y}.\dfrac{dy}{dx}-\dfrac{2020}{x+y}.\dfrac{dy}{dx}=\dfrac{2020}{x+y}-\dfrac{1009}{x}}$

$\mathrm{\Rightarrow(\dfrac{1011}{y}-\dfrac{2020}{x+y}).\dfrac{dy}{dx}=\dfrac{2020}{x+y}-\dfrac{1009}{x}}$

$\mathrm{\Rightarrow\dfrac{1011x+1011y-2020y}{y(x+y)}.\dfrac{dy}{dx}=\dfrac{2020x-1009x-1009y}{x(x+y)}}$

$\mathrm{\Rightarrow\dfrac{1011x-1009y}{y(x+y)}.\dfrac{dy}{dx}=\dfrac{1011x-1009y}{x(x+y)}}$

$\mathrm{\Rightarrow\dfrac{dy}{dx}=\dfrac{y}{x}}$

This is the required derivative.

Read more on Brainly.in

If $x^{1009}y^{1011}=(x+y)^{2020}$, then $\frac{dy}{dx}=$

- https://brainly.in/question/21773191