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

Open in App

Solution

The given curve is $y = x^{3} - x + 1$

$∴ \frac{d y}{d x} = 3 x^{2} - 1$

The slope of the tangent to the curve at $x = 2$ is given by,

${(\frac{d y}{d x})}_{x = 2} = {(3 x^{2} - 1)}_{x = 2}^{2} = 3 (2)^{2} - 1 = 11$

Suggest Corrections

Similar questions

Find the slope of the tangent to the curve y = x³ − 3x + 2 at the point whose x-coordinate is 3.

y = 3 x^{2} - 6

x -

2.

y = x^{3} - x + 1

1

Join BYJU'S Learning Program