`y = x^2 ln(x), (1,0)` Find an equation of the tangent line to the curve at the given point.

Textbook Question

Chapter 3, 3.6 - Problem 34 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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The equation of the tangent line to the curve `y = x^2*ln x` , at the point (1,0) is the following, such that:

`f(x) - f(x_0) = f'(x_0)(x - x_0)`

You need to put` f(x) = y, f(x_0) = 0, x_0 = 1`

You need to evaluate f'(1), hence, you need to find the derivative of the function, using the product rule, such that:

`f'(x) = (x^2)'*ln x + x^2*(ln x)'`

`f'(x) = 2x*ln x + x^2*(1/x)`

`f'(x) = 2x*ln x + x`

Replacing 1 for x, yields:

`f'(1) = 2*ln 1 + 1`

`f'(1) = 2*0 + 1 => f'(1) = 1`

Replacing the values in the equation of tangent line, yields:

`y - 0 = 1*(x - 1)`

`y = x - 1`

Hence, evaluating the equation of the tangent line to the given curve, at the point (1,0), yields `y = x - 1.`

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