[tex] \large \bold{SOLUTION:} [/tex]
Given: [tex] y = (x^2 - 1)^3 + 2x + 3 [/tex]
Required: Slope of the tangent line at [tex]x=2[/tex]
We already know that for a line [tex] y = mx + b, m [/tex] is the slope and [tex]b[/tex] is its y-intercept. To find the slope [tex] m [/tex] of the line tangent to the curve at a given point, we need to differentiate the equation of the curve.
Evaluating [tex]\dfrac{dy}{dx},[/tex] we have
[tex] \qquad \begin{aligned} y &= (x^2 - 1)^3 + 2x + 3 \\ \frac{dy}{dx} &= \frac{d}{dx}\big[(x^2 - 1)^3 + 2x + 3\big] \\ \frac{dy}{dx} &= 3(x^2 - 1)^2 (2x) + 2 \\ \frac{dy}{dx} &= 6x(x^2 - 1)^2 + 2 \end{aligned} [/tex]
Now substitute [tex]x = 2[/tex] to find the slope.
[tex] \qquad \dfrac{dy}{dx}\Big |_{x = 2} = 6(2)(2^2 - 1)^2 + 2 = \boxed{110} [/tex]
Thus, the slope of the line tangent to the curve [tex] y = (x^2 - 1)^3 + 2x + 3 [/tex] at [tex] x = 2 [/tex] is [tex] 110 [/tex].
[tex] \green{\mathfrak{\#CarryOnLearning}} [/tex]
