rules of derivatives
TRANSCRIPT
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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
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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
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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:
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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 .
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 .
22. Derivative of Arctangent:
Explanation
Let . We can get rid of the inverse trig function by rewriting this
as .
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Next, differentiate implicitly:
We must now replace "cosy" with some term involving x.
Since , the triangle at left is formed. The
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
Let . We can get rid of the inverse trig function by rewriting this
as .
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Next, differentiate implicitly:
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
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Then there is a number csuch that a