Question:
Find the slope of the tangent to curve $y=x^{3}-x+1$ at the point whose $x$-coordinate is 2 .
Solution:
The given curve is $y=x^{3}-x+1$.
$\therefore \frac{d y}{d x}=3 x^{2}-1$
The slope of the tangent to a curve at $\left(x_{0}, y_{0}\right)$ is $\left.\frac{d y}{d x}\right]_{\left(x_{0}, y_{0}\right)}$.
It is given that $x_{0}=2$.
Hence, the slope of the tangent at the point where the x-coordinate is 2 is given by,
$\left.\left.\frac{d y}{d x}\right]_{x=2}=3 x^{2}-1\right]_{x=2}=3(2)^{2}-1=12-1=11$
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