Evaluate Derivative at a point
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)=3x2−4x+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)=x3−2x2+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)=3x2−4x+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)2−4(−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)2−4(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)2−4(1)+1=3−4+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)−(x3−5)(4x)(x2)2.
Question 2
Find the gradient of $f\left(x\right)=16x^{-3}$f(x)=16x−3 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).