Browse Questions

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

$\begin{array}{1 1} 24 \\ -24 \\ 20 \\ -20 \end{array}$

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}$=slope of the tangent to $y=f(x)$ at point $P$.
Step 1:
Given : $y=x^3-3x+2$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}$$=3x^2-3 Step 2: \large\frac{dy}{dx} at x=3, \large\frac{dy}{dx}_{(x=3)}=$$3(3)^2-3$
$\qquad\quad\;\;=24$
Hence the required slope of the curve is $24$.