We just looked at our first of many rules for differentiation (the process of finding the derivative).
For a function $f(x)=ax^n$f(x)=axn, the derivative $f'(x)=nax^{n-1}$f′(x)=naxn−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 $x^n$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 |
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.
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)=3x3−3x2, then $f'(x)=9x^2-6x$f′(x)=9x2−6x
c) $f(x)=6x^{-3}-2x+\sqrt{x}$f(x)=6x−3−2x+√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)=6x−3−2x+x12 and so then the derivative $f'(x)=-18x^{-4}-2+\frac{1}{2}x^{-\frac{1}{2}}$f′(x)=−18x−4−2+12x−12
Differentiate $y=2x^3-3x^2-4x+13$y=2x3−3x2−4x+13.
Differentiate $y=7ax^7-2bx^3$y=7ax7−2bx3, where $a$a and $b$b are constants.