Differentiation

New Zealand

Level 7 - NCEA Level 2

Lesson

From all of our work on rates of change we've investigated a lot of ideas, so let's summarise them below.

- For any smoothly varying function, $f(x)$
`f`(`x`), we can find the gradient at any point on the curve. - To find the gradient at a point on the curve we can draw a tangent and find its gradient at that point or we can use the limiting chord process
- If we find the gradient at many points on the curve $f(x)$
`f`(`x`), we can use all these gradients to determine the equation of a new function, the gradient function.

If our original function is linear and of the form $f(x)=mx+c$`f`(`x`)=`m``x`+`c` we know this graph has the same gradient at every point.

If we take $f(x)=x+2$`f`(`x`)=`x`+2, we can see that from the graph or the equation that the gradient is $1$1 everywhere and hence we can represent this gradient function as a graph or table or table of values.

If our original function is a quadratic, we know that the gradient at any point on the curve is different and always changing.

Let's look at the function $f\left(x\right)=x^2$`f`(`x`)=`x`2. Drag the point along the function to find the equation of the tangent line at any point.

If we read off the gradient of the tangent line at different values of $x$`x`, we get this table.

$x$x |
$0$0 | $1$1 | $2$2 | $3$3 |
---|---|---|---|---|

Gradient | $0$0 | $2$2 | $4$4 | $6$6 |

Now let's plot the gradient on the $y$`y`-axis.

If we join the points together, we get the function $y=2x$`y`=2`x`. So we can see that the gradient function of this quadratic is linear. Later on, we will see that this is true for all quadratics.

If our original function was a cubic function, what would our gradient function look like?

We need to be flexible and work back from our gradient function to get an idea of what our original function would have looked like.

The table of values below represent the gradients of various points on the curve $f(x)$`f`(`x`).

(a) Determine the equation of the gradient function.

Think: We can see by looking at the pattern in the gradient values that the gradient function that fits these values will be linear.

Do: Gradient function $=2x-4$=2`x`−4

(b) Which of the following graphs represent the original function?

Think: If our gradient function is linear we know our original function must have been quadratic.

From our table of values we can see that the gradient was $0$0 or flat when $x=2$`x`=2, so we know we must have a turning point here. We can also see from our table of values that for $x<2$`x`<2 our gradients were negative sloping, while for $x>2$`x`>2 our gradients are positive sloping, so we can imagine a minimum turning point.

Do: From the graphs we've been given this means graphs A and D are possible representations of the original function.

Consider the function $f\left(x\right)=-5x+8$`f`(`x`)=−5`x`+8 drawn here.

Which of the following graphs represent $f'\left(x\right)$`f`′(`x`)?

Loading Graph...

- Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...D

Consider the function $y=x^2-1$`y`=`x`2−1 drawn here.

Which of the following graphs represent $y'$`y`′?

Loading Graph...

- Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...D

Consider the gradient function $f'\left(x\right)$`f`′(`x`) drawn here. Which of the following graphs are possible representations of the original function $f\left(x\right)$`f`(`x`)?

Select the two that apply.

Loading Graph...

- Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...D

Apply differentiation and anti-differentiation techniques to polynomials

Apply calculus methods in solving problems