Answer
$y=x-1$
Work Step by Step
Let $f(x)=x^{2}\ln(x)$
The equation of the tangent line to the graph of $f$ at the point $(a,f(a))$ is given by:
$y=f(a)+f'(a)(x-a)$
Since the given point is $(1,0)$ it follows that $a=1$ and $f(1)=0$.
So the equation of the tangent line to the graph of the given function at $(1,0)$ is:
$y=0+f'(1)(x-1)$
Using the product rules for derivatives, the first derivative of $f(x)=x^{2}\ln(x)$ is:
$f'(x)=(x^{2})'\ln{(x)}+x^{2}(\ln{(x)})'$
Using the power rules for derivatives, the first derivative of $x^{2}$is $2x$ so:
$f'(x)=2x\ln{(x)}+x^{2}(\ln{(x)})'$
The first derivative of the natural logarithm is $\frac{1}{x}$ so:
$f'(x)=2x\ln{(x)}+x^{2}\cdot \frac{1}{x}$
Substitute $x=1$ into the equation:
$f'(1)=2\cdot 1\cdot \ln{(1)}+1^{2}\cdot \frac{1}{1}$
Simplify the right side it follows:
$f'(1)=1$
Substitute $f'(1)=1$ into the equation $y=0+f'(1)(x-1)$ it follows:
$y=0+1(x-1)$
Simplify the right side of the equation:
$y=x-1$
Therefore, the equation of the tangent line is:
$y=x-1$
