Rules of Derivatives

8/13/2019 Rules of Derivatives

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Project

In

CALCULUS

Submitted by:

Domingo Reignnel A.

Silla John Immanuel S.

Submitted to:

Miss Leah C. Pira

Date Given: September 12, 2013

Date Submitted: September 26, 2013


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1. Constant Rule

The derivative of a constant is zero; that is, for a constant c:

Problem:

Find the derivative of the function .

Solution

2. Constant Multiple Rule

The derivative of a constant multiplied by a function is the constant multiplied by the

derivative of the original function:

.

Problem.

Solution
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3. Multiple Rules

The derivative of f (axe) is given by

Find the derivative of 3sin (3xs).

Solution

4. Product Rule

The derivative of the product of two functions is NOT the product of the functions'

derivatives; rather, it is described by the equation below:

.

Find the derivative of .

Solution
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5. Quotient Rule

The derivative of the quotient of two functions is NOT the quotient of the functions'

derivatives; rather, it is described by the equation below:

The derivative of

Solution

6. Chain Rule

The chain rule is used to differentiate composite functions. As such, it is a vital tool for

differentiating most functions of a certain complexity.It states:

.The Derivative of

Solution
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7. Power Rule

The power rule is one of several rules used to calculate the derivative of a function. The

power rule states that for every natural number n,the derivative.

Example: What is x

3

?

An = nxn-1

x3=3x

3-1=3x

2

8. Constant power rule

If you raise x to any CONSTANT power, you find the derivative by multiplying x raised to

one less than that power by the power itself. This is the power rule. In equations, if youhave f(x) = an, where n is constant with respect to x, then

df

= f'(x) =

Dx

n xn-1

Examples:

The derivative of F(x) = x5

is equals to f(x) = 5x4

The power rule applies even when the power is not a whole number. The power

can be anything as long as it's constant.

9. The Constant Multiple Rule


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The constant multiple rule says that the derivative of a constant values times a function

is the constant times derivative of the function.

y = 2x4is equals to

10. Sum Rule

The Sum Rule tells us that the derivative of a sum functions, is the sum of the

derivatives.

Differentiate 5x2+ 4x+ 7

11. Difference Rule

The Difference Rule tells us that the derivative of a difference of functions is thedifference of the derivatives.

If fand gare both differentiable, then

12. Sum, Difference, Constant Multiplication And Power Rules

Example: What is (5z2+ z

3- 7z

4) ?

Using the Power Rule:


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z2= 2z

z3= 3z2

z4= 4z3

And so: (5z2+ z3- 7z4) = 5 2z + 3z2- 7 4z3= 10z + 3z2- 28z3

13. Reciprocal Rule

Example: What is (1/x) ?

The Reciprocal Rule says:

the derivative of 1/f = -f/f2

With f(x)= x, we know that f(x) = 1

So:

the derivative of 1/x = -1/x2

DERIVATIVES OF BASIC TRIG FUNCTIONS:

Important note: these derivatives are true only when the angle x is expressed in radians.

This is because thelimit rules we use to evaluate the limits hold only when x is

expressed in degrees.

14. Derivative of Sine:

Explanation

We evaluate the derivative of sine using the definition of derivative:

= by the definition of derivative

=using thetrigonometric identity for

sin(x+y)

=factoring out sin(x) from the terms

containing it.
http://www.math.brown.edu/UTRA/limits.htmlhttp://www.math.brown.edu/UTRA/trigderivs.htmlhttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/trigderivs.htmlhttp://www.math.brown.edu/UTRA/limits.html


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=

separating the rule and pulling the

sin(x) and cos(x) terms outside the

limits

=

by evaluating thetrig limits;xmust

be in radians for these trig limits to

hold

= cosx simplifying the expression

15. Derivative of Cosine:

We could evaluate the derivative of cosine from the definition of derivative,

but it's much easier if we simply use some trig identities and the rule wejust derived for derivative of sine:

Since

cosx =by theco-function identities

= ,

as we're doing the same thing to each side of the

equation,

= by the derivative of sine

= by theco-function identities

16. Derivative of Tangent:

Explanation

We can evaluate the derivative of tangent using thequotient rule and the

derivatives for sine and cosine that we just developed:

Since tan(x) =,

(seequotient trig identities)

=,

since we've done the same thing to each

side of the equation
http://www.math.brown.edu/UTRA/trigderivs.htmlhttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/trigderivs.htmlhttp://www.math.brown.edu/UTRA/derivrules.htmlhttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/derivrules.htmlhttp://www.math.brown.edu/UTRA/trigderivs.htmlhttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/trigderivs.html


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= by thequotient rule

=by evaluating the derivatives for sin and

cos

= by simplifying

= by thePythagorean identity

= sec (x) by thedefinition of secant

DERIVATIVES OF RECIPROCAL TRIG FUNCTIONS:

To find the derivatives of the reciprocal trig functions, we'll simply use the quotient rulewith their definitions in terms of the basic trig functions. Again, these derivatives are true

only when the angle x is expressed in radians.

17. Derivative of Cosecant:

Explanation

= by definition ofcosecant

= by theproduct rule

= by evaluating the derivatives of 1 and sin

= by simplifying and rearranging terms

= by the definitions ofcosecant and cotangent

18. Derivative of Secant:

Explanation
http://www.math.brown.edu/UTRA/derivrules.htmlhttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/trig.html#functionshttp://www.math.brown.edu/UTRA/trigderivs.htmlhttp://www.math.brown.edu/UTRA/trig.html#functionshttp://www.math.brown.edu/UTRA/derivrules.htmlhttp://www.math.brown.edu/UTRA/trig.html#functionshttp://www.math.brown.edu/UTRA/trigderivs.htmlhttp://www.math.brown.edu/UTRA/trigderivs.htmlhttp://www.math.brown.edu/UTRA/trig.html#functionshttp://www.math.brown.edu/UTRA/derivrules.htmlhttp://www.math.brown.edu/UTRA/trig.html#functionshttp://www.math.brown.edu/UTRA/trigderivs.htmlhttp://www.math.brown.edu/UTRA/trig.html#functionshttp://www.math.brown.edu/UTRA/trig.html#identitieshttp://www.math.brown.edu/UTRA/derivrules.html


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= by the definition of secant

= by the product rule

= by evaluating the derivatives of 1 and cos

= by simplifying and rearranging terms

= by the definitions of secant and tangent

19. Derivative of Cotangent:

Explanation

= by the definition of cotangent

= by the product rule

= by evaluating the derivatives of 1 and tan

= by rewriting terms with identities

= by rewriting fractional division as multiplication

= by cancelling out the cos2(x) terms

=csc (x) by the definition of cosecant

DERIVATIVES OF INVERSE TRIG FUNCTIONS:
http://www.math.brown.edu/UTRA/trigderivs.htmlhttp://www.math.brown.edu/UTRA/trigderivs.html


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Again, these derivatives are true only when the angle x is expressed in radians. We'll

follow the same general strategy for calculating each of these derivatives. First, we'll

rewrite the function to remove the inverse expression. We'll then differentiate implicitly,

and we'll finish off by using trig to rewrite all of each derivative in terms of x.

20. Derivative of Arcsine:

Explanation

Let . We can get rid of the inverse trig function by rewriting this

as .

Next, differentiate implicitly:

We must now replace "cosy" with some term

involving x.

Since , the triangle at left is formed. The

bottom leg is found using the Pythagorean

theorem. Using this triangle, we can see

that .

Substituting this into the equation for , we find that


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21. Derivative of Arccosine:

Explanation

Let . We can get rid of the inverse trig function by rewriting thisas .


We must now replace "cosy" with some term

involving x.


bottom leg is found using the Pythagorean theorem.

Using this triangle, we can see that .

Substituting this into the equation for , we find that .

22. Derivative of Arctangent:

Explanation


as .
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We must now replace "cosy" with some term involving x.


hypotenuse is found using the Pythagorean theorem.

Using this triangle, we can see that .

Substituting this into the equation for , we find that .

Of course, each of the reciprocal trig functionscosecant, secant, and cotangentalso

has a corresponding inverse function. Here, we evaluate the derivatives of arc

cosecant, arc secant, and arc cotangent, using the same methods.

23. Derivative of Arc cosecant:

24. Derivative of Arc secant:

Explanation


as .


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We must now replace "(secy)(tany)" with some term involving x.

Since , the triangle at left is formed.

The leg on the right side is found using the

Pythagorean theorem. Using this triangle, we

can see that .

Substituting this into the equation for , we find

that .

25. Derivative of Arc cotangent:

If reciprocal ratios are inputs of inverse functions where the functions have reciprocal

relationships, the resulting angels will be the same!

This fact can be used to rewrite problems and make them easier to solve.

26. The Derivative of the Natural Logarithm Function

If f(x) = in x, then


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f '(x) = 1/x

Example: Find the derivative of

f(x) = in(3x - 4)

Solution

We use the chain rule. We have

(3x - 4)' = 3

and (in u)' = 1/u

Putting this together gives

f '(x) = (3)(1/u)

3

= 3x - 4

27. Exponentials and With Other Bases

Let a> 0 then, a x= exin a

Examples

Find the derivative of

f (x) = 2x

Solution: We write 2x= e

x in 2

Now use the chain rule f '(x) = (exin 2)(in 2) = 2xin 2

28. Logs With Other Bases

We define logarithms with other bases by the change of base formula.


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Definition

in x

logax=

in a

Example: f(x) = log4x

We use the formula

in x

f(x) =

in 4

so that

1

f '(x) =

x in 4

29. Derivative of a Square root

The derivative of the square root of a function is equal to the derivative of the radicand

divided by the double of the root.

Example:

=

30. Exponential Functions

Well start off by looking at the exponential function,


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We want to differentiate this. The power rule that we looked at a couple of sections ago

wont work as that required the exponent to be a fixed number and the base to be a

variable. That is exactly the opposite from what weve got with this function. So, were

going to have to start with the definition of the derivative.

Now, the is not affected by the limit since it doesnt have anyhs in it and so is a

constant as far as the limit is concerned. We can therefore factor this out of the limit.

This gives,

Now lets notice that the limit weve got above is exactly the definition of the derivative

of at , i.e. .

Therefore, the derivative becomes,

31. Logarithm Functions

If f(x)and g(x)are inverses of each other then,

if we have and then,


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The last step just uses the fact that the two functions are inverses of each other.

Putting this all together gives,

Note that we need to require that since this is required for the logarithm

and so must also be required for its derivative. It can also be shown that,

Using this all we need to avoid is .

32. Implicit Differentiation

Example: Find for .

Solution :

This is the simple way of doing the problem. Just solve for yto get the function in the

form that were used to dealing with and then differentiate.

So, thats easy enough to do. However, there are some functions for which this cant be

done. Thats where the second solution technique comes into play.

33. Increasing and Decreasing Functions

Let f be a differentiable function on the interval (a,b) then

1. If f '(x) < 0 for x in (a,b), then f is decreasing there.

2. If f '(x) > 0 for x in (a,b), then f is increasing there.

3. If f '(x) = 0 for x in (a,b), then f is constant.


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Example:

Determine the values of x where the function

f(x) = 2x

3

+ 3x

2

- 12x + 7

Solution

We first take the derivative

f '(x) = 6x2 + 6x - 12

To determine where the derivative is positive and where it is negative, find the roots.

Factor to get

6(x2 + x - 2) = 6(x - 1)(x + 2)

Hence the change in sign can occur when

x = 1 and x = -2

Now create some test values

The derivative is positive outside of [-2,1] and is negative inside of [-2,1]. We can

conclude that f is increasing outside of [-2,1] and decreasing inside of[-2,1].

34. Rolles Theorem

Suppose f(x) is a function that satisfies all of the following.

1. f(x) is continuous on the closed interval [a,b].

2. f(x) is differentiable on the open interval (a,b).

3. f(a) = f(b)

x f '(x)

-3 24

0 -12

2 24


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Then there is a number csuch that a

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