Question

Find the slope of the tangent to curve y = x 3 − x + 1 at the point whose x-coordinate is 2.

Answer 1

solution :- y = x³ - x + 1
we know, slope of tangent = value of 1st derivative of curve.
in mathematically, if f(x) is function.
then, slope of tangent of f(x) at (x = a)= $\frac{dy}{dx}|_{x=a}$
differentiate y with respect to x
$\frac{dy}{dx}=3x^{3-1}-x^{1-1}+0\\\\\frac{dy}{dx}|_{x=2}=3x^2-1\\\\\frac{dy}{dx}|_{x=2}=\text{slope of tangent} = 3\times4-1=11$

slope of tangent = 11