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India
Class XI

Power Rule (x^n)

Lesson

Functions of the type $x^n$xn are called power functions, and the power rule is a shortcut way to find the derivative without having to use first principles every time. 

But, we will use first principles to determine the rule.  

$f(x)$f(x) $=$= $x^n$xn
$f(x+h)$f(x+h) $=$= $(x+h)^n$(x+h)n
$f'(x)$f(x) $=$= $\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$limh0(f(x+h)f(x)h)
$f'(x)$f(x) $=$= $\lim_{h\to0}\left(\frac{\left(x+h\right)^n-x^n}{h}\right)$limh0((x+h)nxnh)

 

From here we need to work through some pretty intense algebra, but it distills down to an understanding of the expansion of $x^n-a^n$xnan

$x^n-a^n=(x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+a^3x^{n-4}+......+a^{n-1})$xnan=(xa)(xn1+axn2+a2xn3+a3xn4+......+an1)

Using this expansion we can also determine the expansion for $(x+h)^n-x^n$(x+h)nxn

$(x+h)^n-x^n==[(x+h)-x][(x+h)^{n-1}+x(x+h)^{n-2}+x^2(x+h)^{n-3}+x^3(x+h)^{n-4}+......+x^{n-1}]$(x+h)nxn==[(x+h)x][(x+h)n1+x(x+h)n2+x2(x+h)n3+x3(x+h)n4+......+xn1]

$f'(x)$f(x) $=$=

$\lim_{h\to0}\left(\frac{h\left[\left(x+h\right)^{n-1}+x\left(x+h\right)^{n-2}+x^2\left(x+h\right)^{n-3}+x^3\left(x+h\right)^{n-4}+...x^{n-1}\right]}{h}\right)$limh0(h[(x+h)n1+x(x+h)n2+x2(x+h)n3+x3(x+h)n4+...xn1]h)

$f'(x)$f(x) $=$=

$\lim_{h\to0}\left(\left(x+h\right)^{n-1}+x\left(x+h\right)^{n-2}+x^2\left(x+h\right)^{n-3}+x^3\left(x+h\right)^{n-4}+...x^{n-1}\right)$limh0((x+h)n1+x(x+h)n2+x2(x+h)n3+x3(x+h)n4+...xn1)

$f'(x)$f(x) $=$=

$x^{n-1}+x\left(x\right)^{n-2}+x^2\left(x\right)^{n-3}+x^3\left(x\right)^{n-4}+...x^{n-1}$xn1+x(x)n2+x2(x)n3+x3(x)n4+...xn1

$f'(x)$f(x) $=$=

$x^{n-1}+\frac{x\left(x\right)^n}{x^2}+\frac{x^2\left(x\right)^n}{x^3}+\frac{x^3\left(x\right)^n}{x^4}+...x^{n-1}$xn1+x(x)nx2+x2(x)nx3+x3(x)nx4+...xn1

$f'(x)$f(x) $=$=

$x^{n-1}+\frac{x^n}{^x}+\frac{x^n}{^x}+\frac{x^n}{^x}+...x^{n-1}$xn1+xnx+xnx+xnx+...xn1

$f'(x)$f(x) $=$=

$x^{n-1}+x^{n-1}+x^{n-1}+x^{n-1}+...x^{n-1}$xn1+xn1+xn1+xn1+...xn1

$f'(x)$f(x) $=$= $nx^{n-1}$nxn1

Now of course, that was a lot of algebraic gymnastics, check with your teacher you may or may not need to follow that entire proof.  For those that don't here is the key message. 

That for functions of the form $f(x)=x^n$f(x)=xn the derivative is $f'(x)=nx^{n-1}$f(x)=nxn1 and this is called the power rule. 

Examples

Find the derivatives of

a)$f(x)=x^2$f(x)=x2, $f'(x)=2x$f(x)=2x

b) $g(m)=m^4$g(m)=m4, so $g'(m)=4m^3$g(m)=4m3

c) $h(t)=t^{\frac{3}{2}}$h(t)=t32, so $h'(t)=\frac{3}{2}t^{\frac{1}{2}}$h(t)=32t12

Be careful with these fractional ones - they are a common source of student errors. 

d) $g(x)=\frac{1}{x^4}$g(x)=1x4, firstly we need the function in power form, so we convert it and get that $g(x)=x^{-4}$g(x)=x4.  Now we can use the power rule and see that $g'(x)=-4x^{-5}$g(x)=4x5   Remember that $-4-1=-5$41=5.  A common mistake here is to arrive at a power of $-3$3.  

 

Power Rule for $x^n$xn!

For a function $f(x)=x^n$f(x)=xn, the derivative $f'(x)=nx^{n-1}$f(x)=nxn1

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

 

 

Example

Question 1

Consider the function $f\left(x\right)=x^n$f(x)=xn, where $n$n is any positive integer.

  1. Complete the following table. (Use the ^ key to create powers).

    $f\left(x\right)$f(x) $f'\left(x\right)$f(x)
    $x$x $\editable{}$
    $x^2$x2 $\editable{}$
    $x^3$x3 $\editable{}$
    $x^4$x4 $\editable{}$
    $x^n$xn $\editable{}$

Question 2

Consider the function $y=\frac{1}{x^2}$y=1x2.

  1. Rewrite the function in negative index form.

  2. Determine the derivative of $y=\frac{1}{x^2}$y=1x2.

Question 3

Determine the derivative of $y=x^{\frac{6}{5}}$y=x65.

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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