New Zealand
Level 7 - NCEA Level 2

# 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 $ax^n$axn

For a function $f(x)=ax^n$f(x)=axn, the derivative $f'(x)=nax^{n-1}$f(x)=naxn1

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

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

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)=2x and $g(x)=x^3$g(x)=x3 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)=x3+2x

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)=2x then $h'(x)=2$h(x)=2

for $g(x)=x^3$g(x)=x3 then $g'(x)=3x^2$g(x)=3x2

and finally for $f(x)=x^3+2x$f(x)=x3+2x

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)=x3+2x

$f(x+h)=(x+h)^3+2(x+h)$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)=h3+3h2x+3hx2+x3+2x+2h

$f(x+h)=x^3+2x+h^3+3h^2x+3hx^2+2h$f(x+h)=x3+2x+h3+3h2x+3hx2+2h

$f(x+h)=[h^3+3h^2x+3hx^2+2h]+[x^3+2x]$f(x+h)=[h3+3h2x+3hx2+2h]+[x3+2x]

 $\frac{dy}{dx}$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 $x^n$xn we saw before, or $ax^n$axn that we are exploring now.

#### Examples

Find the derivative of the following,

a) $f(x)=4x^2+3x+2$f(x)=4x2+3x+2, then $f'(x)=8x+3$f(x)=8x+3   (remember that the derivative of a constant term is $0$0)

b) $f(x)=3x^3-3x^2$f(x)=3x33x2,  then $f'(x)=9x^2-6x$f(x)=9x26x

c) $f(x)=6x^{-3}-2x+\sqrt{x}$f(x)=6x32x+x.  Firstly we need to turn the $\sqrt{x}$x into a power.  $\sqrt{x}=x^{\frac{1}{2}}$x=x12 So $f(x)=6x^{-3}-2x+x^{\frac{1}{2}}$f(x)=6x32x+x12 and so then the derivative $f'(x)=-18x^{-4}-2+\frac{1}{2}x^{-\frac{1}{2}}$f(x)=18x42+12x12

#### Worked Examples:

##### question 1

Differentiate $y=2x^3-3x^2-4x+13$y=2x33x24x+13.

##### question 2

Differentiate $y=7ax^7-2bx^3$y=7ax72bx3, where $a$a and $b$b are constants.

### Outcomes

#### M7-10

Apply differentiation and anti-differentiation techniques to polynomials

#### 91262

Apply calculus methods in solving problems