Question

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

Answer 1

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.