If y-x lnx- then dy dx- 1 - lnx-1-lnx- - - - -none of these-

The correct option is A 1 + lnx Here u=x, v=ln x Differentiating both sides w.r.t. 'x', dy dx=d[x ln x] dx Using product rule = xd( ln x) dx+ln xd(x) dx=x1 x+l ...

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Standard XII

Physics

Basic Differentiation Rule

If y=x lnx, t...

Question

If y=x lnx, then $\frac{d y}{d x}$ =?

1 + lnx

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none of these

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Solution

The correct option is A
1 + lnx

Here u=x, v=ln x
Differentiating both sides w.r.t. 'x', $\frac{d y}{d x} = \frac{d [x l n x]}{d x}$
Using product rule = $x \frac{d (l n x)}{d x} + l n x \frac{d (x)}{d x} = x \frac{1}{x} + l n x .1 = 1 + l n x$

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If y=x lnx, then dydx=?

The correct option is A 1 + lnx Here u=x, v=ln x Differentiating both sides w.r.t. 'x', dydx=d[xlnx]dx Using product rule = xd( ln x)dx+ln xd(x)dx=x1x+lnx.1=1+lnx

If y=x lnx, then $\frac{d y}{d x}$ =?

The correct option is A
1 + lnx

Here u=x, v=ln x
Differentiating both sides w.r.t. 'x', $\frac{d y}{d x} = \frac{d [x l n x]}{d x}$
Using product rule = $x \frac{d (l n x)}{d x} + l n x \frac{d (x)}{d x} = x \frac{1}{x} + l n x .1 = 1 + l n x$