Question

Find the slope of the tangent to the curve \[y = {x^3} - x\] at \[x = 2\].

Answer 1

Hint:
Here, we will find the slope of the tangent is the differentiation of the given curve with respect to \[x\]. Then we will substitute the value of \[x = 2\] in the obtained equation to find the required value.

Complete step by step solution:
We are given that the equation of the curve is
\[y = {x^3} - x{\text{ ......eq.(1)}}\]
We know that the slope of the tangent is the differentiation of the given curve with respect to \[x\].
Differentiating the equation (1) with respect to \[x\], we get
\[
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^3} - x} \right) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{x^3} - \dfrac{d}{{dx}}x \\
   \Rightarrow \dfrac{{dy}}{{dx}} = 3{x^2} - 1 \\
 \]
Substituting the value of \[x = 2\] in the above equation, we get
\[
   \Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 2}} = 3{\left( 2 \right)^2} - 1 \\
   \Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 2}} = 3\left( 4 \right) - 1 \\
   \Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 2}} = 12 - 1 \\
   \Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 2}} = 11 \\
 \]

Hence, we have the slope of the tangent to the curve at \[x = 2\] is 11.

Note:
 We are trying to find the rate of change of one variable compared to another. We should note the first-order derivative of an equation at a specified point is the slope of the line. Also, do not substitute the value of \[x\] before differentiate or else the answer will be wrong. If \[{\left. {\dfrac{{dy}}{{dx}}} \right|_{x = {x_0}}} = 0\], this implies that tangent line is parallel to \[x\]-axis.