21N.2.AHL.TZ0.8

METHOD 1

attempts to differentiate implicitly including at least one application of the product rule                   (M1)

$u=xy, v=\ln \left(xy\right), \frac{du}{dx}=x\frac{dy}{dx}+y, \frac{dv}{dx}=\left(x\frac{dy}{dx}+y\right)\frac{1}{xy}$

$\frac{dy}{dx}=1-\left[\frac{xy}{xy}\left(x\frac{dy}{dx}+y\right)+\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\ln \left(xy\right)\right]$                       A1

Note: Award (M1)A1 for implicitly differentiating $y=x\left(1-y \ln \left(xy\right)\right)$ and obtaining $\frac{dy}{dx}=1-\left[\frac{xy}{xy}\left(x\frac{dy}{dx}+y\right)+x\frac{{\displaystyle dy}}{{\displaystyle dx}}\ln \left(xy\right)+y \ln \left(xy\right)\right]$.

 

$\frac{dy}{dx}=1-\left[\left(x\frac{dy}{dx}+y\right)+\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\ln \left(xy\right)\right]$

$\frac{dy}{dx}=1-\left(x\frac{dy}{dx}+y\right)\left(1+\ln \left(xy\right)\right)$                      A1

$\frac{dy}{dx}+\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\left(1+\ln \left(xy\right)\right)=1$                      AG

 

METHOD 2

$y=x-xy \ln x-xy \ln y$

attempts to differentiate implicitly including at least one application of the product rule                   (M1)

$\frac{dy}{dx}=1-\left(\frac{xy}{x}+\left(x\frac{dy}{dx}+y\right)\ln x\right)-\left(\frac{xy}{y}\frac{dy}{dx}+\left(x\frac{dy}{dx}+y\right)\ln y\right)$                      A1

or equivalent to the above, for example

$\frac{dy}{dx}=1-\left(x \ln x\frac{dy}{dx}+\left(1+\ln x\right)y\right)-\left(y \ln y+x\left(\ln y\frac{dy}{dx}+\frac{dy}{dx}\right)\right)$

$\frac{dy}{dx}=1-x\frac{dy}{dx}\left(\ln x+\ln y+1\right)-y\left(\ln x+\ln y+1\right)$                      A1

or equivalent to the above, for example

$\frac{dy}{dx}=1-x\frac{dy}{dx}\left(\ln \left(xy\right)+1\right)-y\left(\ln \left(xy\right)+1\right)$

$\frac{dy}{dx}+\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\left(1+\ln \left(xy\right)\right)=1$                      AG

 

METHOD 3

attempt to differentiate implicitly including at least one application of the product rule             M1

$u=x \ln \left(xy\right), v=y, \frac{du}{dx}=\ln \left(xy\right)+\left(x\frac{dy}{dx}+y\right)\frac{x}{xy}, \frac{dv}{dx}=\frac{dy}{dx}$

$\frac{dy}{dx}=1-\left(x\frac{dy}{dx}\ln \left(xy\right)+y \ln \left(xy\right)+\frac{xy}{xy}\left(x\frac{dy}{dx}+y\right)\right)$                      A1

$\frac{dy}{dx}=1-x\frac{dy}{dx}\left(\ln \left(xy\right)+1\right)-y\left(\ln \left(xy\right)+1\right)$                      A1

$\frac{dy}{dx}+\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\left(1+\ln \left(xy\right)\right)=1$                      AG

 

METHOD 4

lets $w=xy$ and attempts to find $\frac{dy}{dx}$ where $y=x-w \ln w$             M1

$\frac{dy}{dx}=1-\left(\frac{dw}{dx}+\frac{{\displaystyle dw}}{{\displaystyle dx}}\ln w\right) \left(=1-\frac{dw}{dx}\left(1+\ln w\right)\right)$                      A1

$\frac{dw}{dx}=x\frac{dy}{dx}+y$                      A1

$\frac{dy}{dx}=1-\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y+\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\ln \left(xy\right)\right) \left(=1-\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\left(1+\ln \left(xy\right)\right)\right)$

$\frac{dy}{dx}+\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\left(1+\ln \left(xy\right)\right)=1$                      AG

 

[3 marks]

METHOD 1

substitutes $x=1$ into $y=x-xy \ln \left(xy\right)$                  (M1)

$y=1-y \ln y\Rightarrow y=1$                       A1

substitutes $x=1$ and their non-zero value of $y$ into $\frac{dy}{dx}+\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\left(1+\ln \left(xy\right)\right)=1$                  (M1)

$2\frac{dy}{dx}=0 \left(\frac{dy}{dx}=0\right)$                       A1

equation of the tangent is $y=1$                       A1

 

METHOD 2

substitutes $x=1$ into $\frac{dy}{dx}+\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\left(1+\ln \left(xy\right)\right)=1$                 (M1)

$\frac{dy}{dx}+\left(\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\left(1+\ln \left(y\right)\right)=1$

EITHER

correctly substitutes $\ln y=\frac{1-y}{y}$ into $\frac{dy}{dx}+\left(\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\left(1+\ln \left(xy\right)\right)=1$                       A1

$\frac{dy}{dx}\left(1+\frac{1}{y}\right)=0\Rightarrow \frac{{\displaystyle dy}}{{\displaystyle dx}}=0 \left(y=1\right)$                       A1

OR

correctly substitutes $y+y \ln y=1$ into $\frac{dy}{dx}+\left(\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\left(1+\ln \left(xy\right)\right)=1$                       A1

$\frac{dy}{dx}\left(2+\ln y\right)=0\Rightarrow \frac{{\displaystyle dy}}{{\displaystyle dx}}=0 \left(y=1\right)$                       A1

THEN

substitutes $x=1$ into $y=x-xy \ln \left(xy\right)$                 (M1)

$y=1-y \ln y\Rightarrow y=1$

equation of the tangent is $y=1$                       A1

 

[5 marks]

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