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

**Solution:**

For a curve y = f(x) containing the point (x_{1},y_{1}) the equation of the tangent line to the curve at (x_{1},y_{1}) is given by

y − y_{1} = f′(x_{1}) (x − x_{1})

The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line.

The given curve is

y = x^{3} - x + 1

Therefore,

dy/dx = d/dx x^{3} - x + 1

= 3x^{2} - 1

Now,

the slope of the tangent at the point where the x-coordinate is 2 is given by,

dy/dx]_{x = 2} = 3x^{2} - 1]_{x = 2}

= 3(2)^{2} - 1

= 12 - 1

= 11

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 3

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

**Summary:**

The slope of the tangent to curve y = x^{3} - x + 1 at the point whose x-coordinate is 2 is 11. The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line

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