Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
  • 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,
Step 2:
$\large\frac{dy}{dx}$ at $x=3$,
Hence the required slope of the curve is $24$.
answered Jul 10, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App