Question

if x1009 y1011=(x=y)2020,then 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.

Answer 2

To find:

Derivative of

${x}^{1009} {y}^{1011} = {(x - y)}^{2020}$

Calculation:

The basic trick to solve this type of question is to apply logarithm on both sides;

$= > \log \bigg \{ {x}^{1009} {y}^{1011} \bigg \} = \log \bigg \{{(x - y)}^{2020} \bigg \}$

$= > \log \bigg \{ {x}^{1009} \bigg \} + \log \bigg \{ {y}^{1011} \bigg \} = \log \bigg \{{(x - y)}^{2020} \bigg \}$

$= > 1009\log \bigg \{ x \bigg \} + 1011 \log \bigg \{ y \bigg \} = 2020\log \bigg \{(x - y)\bigg \}$

Now , differentiation w.r.t x ;

$= > 1009 \dfrac{d\log \bigg \{ x \bigg \}}{dx}+ 1011 \dfrac{ d \log \bigg \{ y \bigg \}}{dx} = 2020 \dfrac{d\log \bigg \{(x - y)\bigg \}}{dx}$

$= > \dfrac{ 1009}{x}+ \dfrac{1011}{y}( \dfrac{dy}{dx}) = \dfrac{2020 }{x - y}(1 - \dfrac{dy}{dx} )$

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

$= > \bigg \{\dfrac{1011}{y} + \dfrac{2020}{(x - y)} \bigg \}( \dfrac{dy}{dx} ) = \dfrac{2020 }{x - y} - \dfrac{1009}{x}$

$= > \bigg \{ \dfrac{1011x + 1009y}{y(x - y)} \bigg \}( \dfrac{dy}{dx} ) = \dfrac{1011x + 1009y}{x(x - y)}$

$= > \dfrac{dy}{dx} = \dfrac{y}{x}$

So, final answer is:

$\boxed{ \sf{\dfrac{dy}{dx} = \dfrac{y}{x}}}$