Differentiation

New Zealand

Level 7 - NCEA Level 2

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`)=3`x`2−4`x`+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`)=`x`3−2`x`2+`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`)=3`x`2−4`x`+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.

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`)=`x`2(7`x`2)−(`x`3−5)(4`x`)(`x`2)2.

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

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

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