Differentiation

NZ Level 7 (NZC) Level 2 (NCEA)

Derivative of a Sum (x^n, ax^n)

Lesson

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

Power Rule for $ax^n$`a``x``n`

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

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

(when $a=1$`a`=1, we get the power rule for the simpler power function $x^n$`x``n`)

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

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.

And this applies to any function, whether it be the $x^n$`x``n` we saw before, or $ax^n$`a``x``n` that we are exploring now.

Find the derivative of the following,

a) $f(x)=4x^2+3x+2$`f`(`x`)=4`x`2+3`x`+2, then $f'(x)=8x+3$`f`′(`x`)=8`x`+3 (remember that the derivative of a constant term is $0$0)

b) $f(x)=3x^3-3x^2$`f`(`x`)=3`x`3−3`x`2, then $f'(x)=9x^2-6x$`f`′(`x`)=9`x`2−6`x`

c) $f(x)=6x^{-3}-2x+\sqrt{x}$`f`(`x`)=6`x`−3−2`x`+√`x`. Firstly we need to turn the $\sqrt{x}$√`x` into a power. $\sqrt{x}=x^{\frac{1}{2}}$√`x`=`x`12 So $f(x)=6x^{-3}-2x+x^{\frac{1}{2}}$`f`(`x`)=6`x`−3−2`x`+`x`12 and so then the derivative $f'(x)=-18x^{-4}-2+\frac{1}{2}x^{-\frac{1}{2}}$`f`′(`x`)=−18`x`−4−2+12`x`−12

Differentiate $y=2x^3-3x^2-4x+13$`y`=2`x`3−3`x`2−4`x`+13.

Differentiate $y=7ax^7-2bx^3$`y`=7`a``x`7−2`b``x`3, where $a$`a` and $b$`b` are constants.

Apply differentiation and anti-differentiation techniques to polynomials

Apply calculus methods in solving problems