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If y=(lnx)/(x)"then"(dy)/(dx) will be:...

If `y=(lnx)/(x)`"then"`(dy)/(dx)` will be:

A

`(1-lnx)/(x)`

B

`(1+lnx)/(x^(2))`

C

`(1-lnx)/(x^(2))`

D

`(lnx-1)/(x^(2))`

Text Solution

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The correct Answer is:
To find the derivative of the function \( y = \frac{\ln x}{x} \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), then the derivative is given by: \[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \] In our case, we can identify: - \( u = \ln x \) - \( v = x \) Now, we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \): 1. **Differentiate \( u = \ln x \)**: \[ \frac{du}{dx} = \frac{1}{x} \] 2. **Differentiate \( v = x \)**: \[ \frac{dv}{dx} = 1 \] Now we can substitute these derivatives into the quotient rule formula: \[ \frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^2} \] This simplifies to: \[ \frac{dy}{dx} = \frac{1 - \ln x}{x^2} \] Thus, the derivative of \( y = \frac{\ln x}{x} \) is: \[ \frac{dy}{dx} = \frac{1 - \ln x}{x^2} \]

To find the derivative of the function \( y = \frac{\ln x}{x} \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), then the derivative is given by: \[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \] In our case, we can identify: - \( u = \ln x \) ...
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