Derivative Rules

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Derivatives The Easy Way

Constant Rule and Power Rule

We have seen the following derivatives:

If f(x) = c, then f '(x) = 0
If f(x) = x, then f '(x) = 1
If f(x) = x², then f '(x) = 2x
If f(x) = x³, then f '(x) = 3x²
If f(x) = x⁴, then f '(x) = 4x³

This leads us the guess the following theorem.

Theorem

      d
            xⁿ = nxⁿ^-1
     dx

Proof:

We have

\(\lim\limits_{h \to 0} \frac{(x + h)^n - x^n}{h} = \lim\limits_{h \to 0} \frac{x^n + nx^{n-1}h + C(n,2)x^{n-2}h^2 + ... + nxh^{n-1} + h^n - x^n}{h} \)

\(= \lim\limits_{h \to 0} \frac{nx^{n-1}h + C(n,2)x^{n-2}h^2 + ... + nxh^{n-1} + h^n} {h} \)

\(= \lim\limits_{h \to 0} \frac{h(nx^{n-1} + C(n,2)x^{n-2}h + ... + nxh^{n} + h^{n-1}} {h} \)

\(= \lim\limits_{h \to 0} \frac{nx^{n-1} + C(n,2)x^{n-2}h + ... + nxh^{n} + h^{n-1} }{1} = nx^{n - 1} \)

Applications

Example

Find the derivatives of the following functions:

f(x) = 4x³ - 2x¹⁰⁰
f(x) = 3x⁵ + 4x⁸ - x + 2
f(x) = (x³ - 2)²

Solution

We use our new derivative rules to find

12x² - 200x⁹⁹
15x³+32x⁷-1
First we FOIL to get

[x⁶- 4x³+ 4] '

Now use the derivative rule for powers

6x⁵ - 12x²

Example:

Find the equation to the tangent line to

y = 3x³ - x + 4

at the point (1,6)

Solution:

        y' = 9x² - 1

at x = 1 this is 8. Using the point-slope equation for the line gives

        y - 6 = 8(x - 1)

or

        y = 8x - 2

Example:

Find the points where the tangent line to

y = x³ - 3x²- 24x + 3

is horizontal.

Solution:

We find

        y' = 3x² - 6x - 24

The tangent line will be horizontal when its slope is zero, that is, the derivative is zero. Setting the derivative equal to zero gives:

        3x² - 6x - 24 = 0

or

        x² - 2x - 8 = 0

or

        (x - 4)(x + 2) = 0

so that

        x = 4    or    x = -2

Derivative of f(x) = sin(x)

Theorem

      d
            sin(x) = cos(x)
     dx

Proof:

\(= \lim\limits_{h \to 0} \frac{sin(x+h) - sin(x)} {h} = \lim\limits_{h \to 0} \frac{sin(x)cos(h) + sin(h)cos(x) - sin(x)} {h} \)

\(= \lim\limits_{h \to 0} [\frac{sin(x)cos(h) - sin(x)} {h} + \frac{sin(h)cos(x)}{h}] = sin(x)\lim\limits_{h \to 0} [\frac{cos(h) - h} {h}] + cos(x)\lim\limits_{h \to 0} [\frac{sin(h)cos(x)} {h}] \)

\( = (sin(x))(0) + (cos(x))(1) = cos(x) \)

d/dx cos(x)

Theorem

       d
             cos x = -sin x
      dx

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