#### To determine

**To evaluate:** Equation of the tangent line of * y* at the point (1, 0)

#### Answer

The equation of tangent line of *y* at (1, 0) is: y=x−1

#### Explanation

**Given that:** y=x2ln(x)

**Formula used:**

1. Product rule: uv'=u'v+uv'

2. Equation of line with slope *m* and passing point (x1,y1) is: y−y1=m(x−x1)

**Calculation:**

For calculating tangent line we have to find slope of the tangent line. We know that the slope of the tangent line is nothing but the derivative of the function at tangent point. So, first we have to calculate y'(1).

Consider,

y=x2ln(x)

Differentiate *y* with respect to x, we get

y'=ddx[x2ln(x)]=ddx(x2)lnx+x2ddx[lnx] (By product rule)=2xlnx+x2×1x=2xlnx+x=x(2lnx+1)

Put x=1 in above, we get

y'(1)=1(2ln1+1) =1(0+1) =1

Therefore, slope of the tangent line at the point (1, 0) is y'(1)=1. So, by formula 2, equation of line with slope y'(1)=1 and passing point (1, 0) is: (y−0)=1(x−1) i.e.

y=x−1

**Conclusion:** Thus, the equation of the tangent of *y* at (1, 0) is: y=x−1