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

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  • 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 $24$.
answered Jul 10, 2013 by sreemathi.v
 

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