# How do you find the slope of the line tangent to the graph of y = x ln x at the point ( 1,0 )?

Nov 14, 2016

$y = x - 1$

#### Explanation:

The slope of the tangent at any particular point is given by the derivative.

We have $y = x \ln x$

Differentiating wrt $x$ using the product rule gives us:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(x\right) \left(\frac{d}{\mathrm{dx}} \ln x\right) + \left(\frac{d}{\mathrm{dx}} x\right) \left(\ln x\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left(x\right) \left(\frac{1}{x}\right) + \left(1\right) \left(\ln x\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \ln x$

So, At$\left(1.0\right) , \frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \ln 1 = 1$

So the required tangent passes through $\left(1.0\right)$ and has slope $1$
Using $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the required equation is;
$y - 0 = 1 \left(x - 1\right)$
$\therefore y = x - 1$