Question

find the slope of the tangent line to the graph of f(x) = 2x ln x at the point (1,0).

Explanation:

Step1: Recall derivative formula

The derivative of a function gives the slope of the tangent line. For $y = 2x\ln x$, use the product - rule $(uv)'=u'v + uv'$, where $u = 2x$ and $v=\ln x$. The derivative of $u = 2x$ is $u'=2$, and the derivative of $v=\ln x$ is $v'=\frac{1}{x}$.

Step2: Apply product - rule

$y'=(2x)'\ln x+2x(\ln x)'=2\ln x + 2x\cdot\frac{1}{x}=2\ln x + 2$.

Step3: Evaluate derivative at the given point

We want to find the slope of the tangent line at the point $(1,0)$. Substitute $x = 1$ into $y'$. When $x = 1$, $y'(1)=2\ln(1)+2$. Since $\ln(1)=0$, then $y'(1)=2$.

Answer:

$2$