Differentiation

New Zealand

Level 7 - NCEA Level 2

Lesson

We just looked a our first of many rules for differentiation (the process of finding the derivative).

Power Rule for $x^n$`x``n`!

For a function $f(x)=x^n$`f`(`x`)=`x``n`, the derivative $f'(x)=nx^{n-1}$`f`′(`x`)=`n``x``n`−1

$n$`n` can be positive or negative, integer or fraction

Adding together terms in varying power forms create other functions, which in turn we can find the derivative of.

For example if we have the functions $h(x)=2x$`h`(`x`)=2`x` and $g(x)=x^3$`g`(`x`)=`x`3 then we can add $h(x)$`h`(`x`) and $g(x)$`g`(`x`) together to create a new function $f(x)=x^2+2x$`f`(`x`)=`x`2+2`x`.

To see what happens with the sum of the function and the resulting gradient function, let's look at 3 parts. Firstly $h(x)$`h`(`x`), then $g(x)$`g`(`x`) and then $f(x)$`f`(`x`).

$h(x)=2x$`h`(`x`)=2`x` then $h'(x)=2$`h`′(`x`)=2

for $g(x)=x^3$`g`(`x`)=`x`3 then $g'(x)=3x^2$`g`′(`x`)=3`x`2

and finally for $f(x)=x^3+2x$`f`(`x`)=`x`3+2`x`

Well, we are not sure yet how to find this so for the last one we will use first principles.

So we need to know $f(x)$`f`(`x`) and $f(x+h)$`f`(`x`+`h`)

$f(x)=x^3+2x$`f`(`x`)=`x`3+2`x`

$f(x+h)=(x+h)^3+2(x+h)$`f`(`x`+`h`)=(`x`+`h`)3+2(`x`+`h`)

$f(x+h)=h^3+3h^2x+3hx^2+x^3+2x+2h$`f`(`x`+`h`)=`h`3+3`h`2`x`+3`h``x`2+`x`3+2`x`+2`h`

$f(x+h)=x^3+2x+h^3+3h^2x+3hx^2+2h$`f`(`x`+`h`)=`x`3+2`x`+`h`3+3`h`2`x`+3`h``x`2+2`h`

$f(x+h)=[h^3+3h^2x+3hx^2+2h]+[x^3+2x]$`f`(`x`+`h`)=[`h`3+3`h`2`x`+3`h``x`2+2`h`]+[`x`3+2`x`]

$\frac{dy}{dx}$dydx |
$=$= | |

$\frac{dy}{dx}$dydx |
$=$= | |

$\frac{dy}{dx}$dydx |
$=$= | |

$\frac{dy}{dx}$dydx |
$=$= | |

$\frac{dy}{dx}$dydx |
$=$= | |

$\frac{dy}{dx}$dydx |
$=$= | $3x^2+2$3x2+2 |

So we can see that the derivative of the sum is the same as the sums of the derivatives.

Sum of Derivatives

Derivative of sum is equal to the sum of the derivatives.

If $f(x)=g(x)\pm h(x)$`f`(`x`)=`g`(`x`)±`h`(`x`) then $f'(x)=g'(x)\pm h'(x)$`f`′(`x`)=`g`′(`x`)±`h`′(`x`)

This means we can apply the power rule to individual terms.

Find the derivative of the following,

a) $f(x)=x^2+x+2$`f`(`x`)=`x`2+`x`+2, then $f'(x)=2x+1$`f`′(`x`)=2`x`+1 (remember that the derivative of a constant term is $0$0)

b) $f(x)=x^3-x^2$`f`(`x`)=`x`3−`x`2, then $f'(x)=3x^2-2x$`f`′(`x`)=3`x`2−2`x`

c) $f(x)=x^{-3}-x+\sqrt{x}$`f`(`x`)=`x`−3−`x`+√`x`. Firstly we need to turn the $\sqrt{x}$√`x` into a power. $\sqrt{x}=x^{\frac{1}{2}}$√`x`=`x`12 So $f(x)=x^{-3}-x+x^{\frac{1}{2}}$`f`(`x`)=`x`−3−`x`+`x`12 and so then the derivative $f'(x)=-3x^{-4}-1+\frac{1}{2}x^{-\frac{1}{2}}$`f`′(`x`)=−3`x`−4−1+12`x`−12

Determine the derivative of $y=x^4+x^5$`y`=`x`4+`x`5.

Find the derivative of $y=x^3+x^{-5}-9$`y`=`x`3+`x`−5−9.

Give your answer with positive indices.

Find the derivative of $y=\frac{x+1}{\sqrt[7]{x}}$`y`=`x`+1^{7}√`x`.

Give your answer with positive indices.

Apply differentiation and anti-differentiation techniques to polynomials

Apply calculus methods in solving problems