Practicing Success
If $y=x \log x$, then which of the following is correct ? |
$x \frac{d y}{d x}-x=y$ $x \frac{d y}{d x}+y=x$ $x \frac{d y}{d x}+x y=0$ $y \frac{d y}{d x}+x=y$ |
$x \frac{d y}{d x}-x=y$ |
$y=x \log x-1$ ......(1) differentiating (1) w.r.t (x) $\frac{d y}{d x} =\frac{d}{d x}(x \log x) \Rightarrow \frac{d y}{d x}=\log x \frac{d x}{d x}+x \frac{d \log x}{d x}$ $\Rightarrow \frac{d y}{d x} =\log x . 1+x \times \frac{1}{x}$ $\frac{d y}{d x} =\log x+1$ ......(2) now multiplying (2) with (x) $\frac{x d y}{d x}=x \log x+x$ Substituting x log x with y from eq (1) $\frac{x d y}{d x}=x+y \Rightarrow x\frac{d y}{d x}-x=y$ |