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## ■Differentiation

### Differentiation of sin(x), cos(x) and tan(x)

sin(x)
It is not difficult to differentiate a function such as x^2 based on the definition. However, we need to be more creative to differentiate a trigonometric function such as sin(x).

Simplify the definition of differentiation for sin(x).

The limit of at h=0 is cos(x) because the limit of sin(x)/x at x=0 is 1.
(See the limit of sin(x)/x earlier in this document.)
Multiply the denominator and numerator of by cos(h)+1 and transform the expression.

When h is positive and small,

< because sin(h)<h.
As h approaches 0, cos(h) approaches 1 and approaches 0.
Thus, becomes 0.

Finally the limit of sin(x) at h=0 is,
lim( )
= lim( ) + lim( )
= cos(x) + 0
= cos(x)

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cos(x)
A similar proof can be used to prove the differentiation of cos(x).

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tan(x)
We can use the quotient rule to prove the differentiation of tan(x).

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### Differentiation of ln(x)

It is important to remember the definition of the natural logarithm base e.

The definition of the differentiation of ln(x) is:

We will transform the expression with the conditions that x>0 and h>0.

Suppose h=t*x and that t approaches 0 as h approaches 0.

Apply logarithm properties.

Calculate the limit.

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### Differentiation of ex

We will transform the expression and let t = eh-1.

Apply logarithm properties.

Calculate with the definition .

We can differentiate ax using the product rule by first transforming ax into exp^(x*ln(a)).

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### Tangent line to y = f(x) and f(x, y) = 0

The differentiation gives the slope of curve. For example, the slope of y = x^2 is d/dx(x^2 )= 2x at the point (x, x^2).
Suppose the tangent line is y = mx+b. The tangent line at (1, 1) is y = 2x+b because the slope is 2. Substitute (1, 1) to y = 2x+b, we find b = -1.

ClassPad has a tanLine function to have tangent easily and a normal function to have a normal line.

When the given function is implicit, ClassPad has an impDiff function to differentiate the implicit function such as Unit circle x^2+y^2 = 1.
When doing impDiff(x^2+y^2 = 1, x, y), you have y' = -x/y. It's a slope of the circle.
The slope at (1/2, √(3)/2) is -√(3)/3, and the tangent line is y = -√(3)/3x-√(3)/3.

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### Differentiation of y = f(x) or f(x, y) = 0 and its graph

The differentiation of y = x^2 is y' = 2x which is a line.

The differentiation of Unit circle x^2+y^2 = 1 is y' = -x/y which are two curves, y = x/√(1-x^2) and y = -x/√(1-x^2).

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### Newton's Method

Newton's method is a method used to find the x-intercept of the function y = f(x). The recurrence equation used is Xn+1=Xn-f(Xn)/f'(Xn).

We find that there are four roots for the graph of f(x) = x^4-3x^2+sin(x).

(You can approximate the roots by pressing on the graph.)

An initial value for Newton's method is assumed to be 2 because the solution of the rightmost is almost 1.6.

This recurrence shows us the solution approaches to 1.618.
The Cobweb diagram gives another way to use Newton's method and the graph.

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### Differentiation presented by parametric equations

The differentiation for parametric equations such as (x(t), y(t)) is,

The following example uses {u(t)=7t+2, v(t)=t^3-12t}.

The second derivative is as follows.

The following example is the first and second derivative for {x(t)=t^2, y(t)=t^4+1}.

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### Slope in Polar form

The slope in Polar form r = f(θ) is as follows.

The next example shows how to calculate the slope at θ=/6 for r = 1+cos(θ).

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