Application of Derivatives — Class 12 Maths Solution

ncert exercise SA NCERT Ex. 6.3, Q.3,Page 211
Question

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

Step-by-step Solution

We have, $y = {x^3} - x + 1$ …(i)

Differentiating (i) w.r.t. $x$, we get $\cfrac{{dy}}{{dx}} = 3{x^2} - 1$

Therefore, Slope of tangent at $x = 2$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{x = 2}} = 3{(2)^2} - 1 = 11$

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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Application of Derivatives. Curated by Sachin Sharma. Free for all students.