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

$\begin{array}{1 1} 11 \\ -11 \\ 21 \\ -21 \end{array}$

## 1 Answer

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}$=slope of the tangent to $y=f(x)$ at point $P$.
Step 1:
Given : $y=x^3-x+1$
Differentiate w.r.t $x$ we get,
$\large\frac{dy}{dx}$$=3x^2-1 Step 2: The slope of the tangent when x=2 \large\frac{dy}{dx}_{(x-2)}$$=3(2)^2-1$
$\qquad\quad=12-1$
$\qquad\quad=11$
Slope of the tangent to the curve is $11$.
answered Jul 10, 2013

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer