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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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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} $

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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 by sreemathi.v
 

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