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India
Class XI

Evaluate Derivative at a point

Lesson

Having found the derivative function, you can evaluate the derivative at a point by substitution in any value of $x$x.

For the following derivative function $f'(x)=3x^2-4x+1$f(x)=3x24x+1 the values of the derivative at the points on the curve where $x=-1,0$x=1,0 and $1$1

If the original function was $f(x)=x^3-2x^2+x+7$f(x)=x32x2+x+7.  Then it's graph would look like this. 

The points we are interested in are at $x=-1,0$x=1,0 and $1$1.  Marked here on the graphs with their tangents.

Now we know that the derivative function is $f'(x)=3x^2-4x+1$f(x)=3x24x+1, so the derivative at $x=-1$x=1 is $f'(-1)$f(1), the derivative at $x=0$x=0 is $f'(0)$f(0) and the derivative at $x=1$x=1, is $f'(1)$f(1).

$f'(-1)=3(-1)^2-4(-1)+1=3+4+1=8$f(1)=3(1)24(1)+1=3+4+1=8   This is pretty steep, and the graph above confirms this.

$f(0)=3(0)^2-4(0)+1=1$f(0)=3(0)24(0)+1=1.  This looks reasonable on the graph above

$f(1)=3(1)^2-4(1)+1=3-4+1=0$f(1)=3(1)24(1)+1=34+1=0.  And we can tell the horizontal tangent on the graph above also. 

 

All we need to be able to evaluate the derivative, is the $x$x coordinate of the point and the gradient function.  

Examples

Question 1

Simplify $f'\left(x\right)=\frac{x^2\left(7x^2\right)-\left(x^3-5\right)\left(4x\right)}{\left(x^2\right)^2}$f(x)=x2(7x2)(x35)(4x)(x2)2.

Question 2

Find the gradient of $f\left(x\right)=16x^{-3}$f(x)=16x3 at $x=2$x=2.

Denote this gradient by $f'\left(2\right)$f(2).

Question 3

By considering the graph of $f\left(x\right)=-6$f(x)=6, find $f'\left(4\right)$f(4).

Outcomes

11.C.LD.1

Derivative introduced as rate of change both as that of distance function and geometrically, intuitive idea of limit. Definition of derivative, relate it to slope of tangent of the curve, derivative of sum, difference, product and quotient of functions. Derivatives of polynomial and trigonometric functions.

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