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India
Class XI

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

11.C.LD.1

Derivative introduced as rate of change both as that of distance function and geometrically, intuitive idea of limit. Definition of derivative, relate it to slope of tangent of the curve, derivative of sum, difference, product and quotient of functions. Derivatives of polynomial and trigonometric functions.

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