- A${{y(x{{\log }_e}y + y)} \over {x(y{{\log }_e}x + x)}}$
- ✓${{y(x{{\log }_e}y - y)} \over {x(y{{\log }_e}x - x)}}$
- C${{x(x{{\log }_e}y - y)} \over {y(y{{\log }_e}x - x)}}$
- D${{x(x{{\log }_e}y + y)} \over {y(y{{\log }_e}x + x)}}$
Differentiating w.r.t. $x$ of $y,$ we get
${\log _e}x\frac{{dy}}{{dx}} + \frac{y}{x} = {\log _e}y + x\frac{1}{y}\frac{{dy}}{{dx}}$
$\therefore \frac{{dy}}{{dx}} = \frac{{y(x{{\log }_e}y - y)}}{{x(y{{\log }_e}x - x)}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f(x) = {\rm{ }}\left\{ {\begin{array}{*{20}{c}}
{\frac{{\left( {{e^x} - 1} \right)^2}}{{\sin {\mkern 1mu} \left( {\frac{x}{k}} \right){\mkern 1mu} \log {\mkern 1mu} \left( {1 + \frac{x}{4}} \right)}}{\mkern 1mu} ,{\mkern 1mu} x \ne 0}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 12{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ,x{\mkern 1mu} {\mkern 1mu} = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} }
\end{array}} \right.$ એ સતત વિધેય હોય તો $k$ ની કિમંત મેળવો.