Question

if y=x log x then dy/dx=?

Answer 1

Answer:

by product rule....

dy/dx = 1.logx + x. 1/x

= 1 + logx

Answer 2

Answer:

$\frac{dy}{dx} = 1 + log(x)$

Step-by-step explanation:

Given, y = x log(x)

To find dy/dx, We'll differentiate the given expression with respect to x.

$\frac{ d}{dx} (y) = \frac{d}{dx} (x log(x) )$

To solve the Right hand side expression further, We'll use Product rule of differentiation which is,

$\frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

So,

$\frac{dy}{dx} = x \frac{d}{dx} ( log(x) ) + log(x) \frac{d}{dx} (x)$

We can derive from the first principle that,

$\frac{d}{dx}( log(x)) = \frac{1}{x} \: \\ \frac{d}{dx} (x) = 1$

(Assuming that, log here refers to Natural logarithm)

We have,

$\frac{dy}{dx} = (x \times \frac{1}{x} )+ ( log(x) \times (1)) \\ \frac{ dy}{dx} = 1 + log(x)$

This is the required answer.