Question Number 25519 by Mahesh Andiboina last updated on 11/Dec/17
Commented by Mahesh Andiboina last updated on 11/Dec/17
$$\mathrm{plz}\:\mathrm{ans}\:\mathrm{both} \\ $$
Answered by sudhanshur last updated on 11/Dec/17
$${e}^{−{xy}} −\mathrm{4}{xy}=\mathrm{0} \\ $$$${e}^{−{xy}} =\mathrm{4}{xy} \\ $$$$−{xy}=\mathrm{ln}\:\left(\mathrm{4}{xy}\right)=\mathrm{ln}\:\mathrm{4}+\mathrm{ln}\:{x}+\mathrm{ln}\:{y} \\ $$$$−\frac{{d}}{{dx}}\left({xy}\right)=\mathrm{0}+\frac{\mathrm{1}}{{x}}+\frac{{d}}{{dy}}\mathrm{ln}\:{y}\centerdot\frac{{dy}}{{dx}} \\ $$$$−{x}\frac{{dy}}{{dx}}−{y}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}} \\ $$$$−\left({x}+\frac{\mathrm{1}}{{y}}\right)\frac{{dy}}{{dx}}=\left(\frac{\mathrm{1}}{{x}}+{y}\right) \\ $$$$−\frac{{xy}+\mathrm{1}}{{y}}\centerdot\frac{{dy}}{{dx}}=\frac{\mathrm{1}+{xy}}{{x}} \\ $$$$\frac{{dy}}{{dx}}=−\frac{{y}}{{x}} \\ $$
Commented by Rasheed.Sindhi last updated on 11/Dec/17
$$\mathrm{Font}\:\mathrm{size}\:\mathrm{too}\:\mathrm{small}! \\ $$
Commented by Mahesh Andiboina last updated on 11/Dec/17
$$\mathrm{sir}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{actually}\:\mathrm{e}^{−\mathrm{xy}} \\ $$
Commented by Mahesh Andiboina last updated on 11/Dec/17
$$\mathrm{kk}\:\mathrm{sir}\:\mathrm{also}\:\mathrm{ans}\:\mathrm{the}\:\mathrm{13}\:\mathrm{th}\:\mathrm{question} \\ $$
Answered by sushmitak last updated on 11/Dec/17
$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}−\mathrm{2}{y}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=\frac{{x}}{{y}} \\ $$$${option}\:\left({a}\right)\:\mathrm{is}\:\mathrm{correct}. \\ $$
Commented by Mahesh Andiboina last updated on 11/Dec/17
$$\mathrm{i}\:\mathrm{didnt}\:\mathrm{understand} \\ $$
Commented by Mahesh Andiboina last updated on 11/Dec/17
$$\mathrm{kk}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{right} \\ $$