05 5polynomials

# The derivative of xn

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If … $f(x) = x^n$

…then … $f'(x) = nx^{n -1}$

• bring the index down and multiply it at the front of the term
• subtract 1 from the index

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Eg if … $f(x) = x^4$

… then … $f'(x) = 4x^3$ .

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If … $f(x) = ax^n$ …. {where a is any constant}

… then … $f'(x) = anx^{n - 1}$
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Eg if … $f(x) = 5x^4$,

… then … $f'(x) = 20x^3$

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If … $f(x) = a$ ….. {where a is any constant}

… then … $f'(x) = 0$ …. {because $a = ax^0 , \;so\; 0 \times a = 0$}
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Eg if .. $f(x) = 5$,

… then … $f'(x) = 0$

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## The derivative of Polynomials

To find the derivative of any polynomial

• find the derivative of each term and add the results

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Example

Find the derivative of … $f(x) = 3x^2 - 7x + 3$

Solution:

… … $f(x) = 3x^2 - 7x^1 + 3x^0$

… … $f'(x) = 2 \big( 3x^{2-1} \big) - 1 \big( 7x^{1-1} \big) + 0 \big( x^{0-1} \big)$

… … $f'(x) = 6x^1 - 7x^0 + 0$

… … $f'(x) = 6x - 7$

Note 1

• Remember that $\dfrac{dy}{dx}$ is an alternate notation for derivative

… … if $y = x^2 + 3x + 4$

… … then $\dfrac{dy}{dx} = 2x + 3$

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Note 2

• These rules work even if the exponent, n is negative or a fraction
• bring the index down and multiply it at the front of the term
• subtract one from the index

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• Recall that

… … $\dfrac{1}{x^3} = x^{-3}$
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… … if $y = x^{-3}$

… … then $\dfrac{dy}{dx} = -3x^{-4} = \dfrac{-3}{x^4}$

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*Recall that

… … $\sqrt {x} = x^{\dfrac{1}{2}}$

… … and $\quad \sqrt [3] {x} = x^{\dfrac{1}{3}}$
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… … if $y = x^{\dfrac{1}{3}}$

… … then $\dfrac{dy}{dx} = \dfrac{1}{3}x^{-\dfrac{2}{3}}$

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## Differentiation on the Calculator

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Select the derivative icon from Math2 tab
• You will need to put x into the denominator of the derivative.
• Then enter the expression to be differentiated between the brackets and press EXE.
• Avoid writing calculator notation when you answer questions.
• for example, write 2x and not 2 • x

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#### Example 1

Find the derivative of .. $y = x^2 - 5x - 24$

Solution:

… … $y=x^2-5x-24$

… … $\dfrac{dy}{dx}=2x-5$

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#### Example 2

Find the derivative of $f(x)=\dfrac{3x^2-2x+1}{x}$

and hence find $f'(1)$

Solution:

… … $f(x) = \dfrac{3x^2-2x+1}{x}$

… … break into separate fractions

… … $f(x) = \dfrac{3x^2}{x} - \dfrac{2x}{x} + \dfrac{1}{x}$

… … $f(x) =3x-2+\dfrac{1}{x}$

… … $f(x) = 3x-2+x^{-1}$

… … therefore the derivative is:

… … $f'(x)=3-0-x^{-2}$

… … $f'(x) = 3-\dfrac{1}{x^2}$

and hence find $f'(1)$

… … $f'(1)=3-\dfrac{1}{1^2}$

… … $f'(1) =2$

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Note 1

On your calculator, entering "| x = 1" after the derivative gives $f'(1)$.

• The "|" symbol is in the Math3 tab.

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Note 2

If you intend to do several calculations with the same function, it is worth defining the function, first.

• Use "f" from the ABC tab of the virtual keyboard, but don't use "x" from the same place.
• Use the variable version of "x".

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Note 3

The calculator will sometimes give the result in a different format from the way you obtain your answer
see answer for Example 2 on right
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• The expand command will split the combined fraction into separate fractions
• The simplify command (same place) will usually do the same thing.
• The combine command (same place) will combine 2 fractions into a single fraction.

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Example 3

Differentiate $y=\sqrt{x}-\dfrac{2}{\sqrt{x}}$

Solution:

… … $y = \sqrt{x} - \dfrac{2}{\sqrt{x}}$

… … $y = x^{\dfrac{1}{2}} - 2x^{-\dfrac{1}{2}}$

… … $\dfrac{dy}{dx} = \dfrac{1}{2} x^ {-\dfrac{1}{2}}-2 \left( -\dfrac{1}{2}x^{-\dfrac{3}{2}} \right)$

… … $\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}+\dfrac{1}{\sqrt{x^3}}$

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Note:

A function can only be differentiated at a point if its graph is continuous and smooth at that point.
Also, we can not differentiate the endpoint of a function.

This is because differentiating is equivalent to finding a tangent to the curve at that point

• If the graph is discontinuous, or an endpoint, or not smooth then the tangent is undefined.

*See here for the conditions of differentiation.

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