Given: [tex]\rm x y=\log y+C[/tex]
On differentiating both sides w.r.t x,
[tex]\[ \begin{array}{l}\\ \displaystyle \rm x y^{\prime}+y=\frac{1}{y} \cdot y^{\prime} \\\\ \\ \displaystyle \rm \Rightarrow \frac{y^{\prime}}{y}-x y^{\prime}=y \\ \\ \\ \displaystyle \rm\Rightarrow y^{\prime}\left(\frac{1}{y}-x\right)=y \\ \\ \\ \displaystyle \rm\Rightarrow y^{\prime}\left(\frac{1-x y}{y}\right)=y \\ \\ \\ \displaystyle \rm\Rightarrow y^{\prime}=\frac{y^{2}}{1-x y} \end{array} \][/tex]
Hence, proved.
Step-by-step explanation:
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Answer 3:-
Given: [tex]\rm x y=\log y+C[/tex]
On differentiating both sides w.r.t x,
[tex]\[ \begin{array}{l}\\ \displaystyle \rm x y^{\prime}+y=\frac{1}{y} \cdot y^{\prime} \\\\ \\ \displaystyle \rm \Rightarrow \frac{y^{\prime}}{y}-x y^{\prime}=y \\ \\ \\ \displaystyle \rm\Rightarrow y^{\prime}\left(\frac{1}{y}-x\right)=y \\ \\ \\ \displaystyle \rm\Rightarrow y^{\prime}\left(\frac{1-x y}{y}\right)=y \\ \\ \\ \displaystyle \rm\Rightarrow y^{\prime}=\frac{y^{2}}{1-x y} \end{array} \][/tex]
Hence, proved.
Step-by-step explanation:
Write a report for the school magazine about a farewell given to the 12th Class Students . Plz write in a proper format .
me bhi mast hu...aap to bhul gye the (˘･_･˘)
do you have any social media app...?
(・_・;)