Differentiation Profit, Cost, and Revenue Linear Programming Integration
Differentiation
Without too much attention to detail and theory, we have six basic ways to differentiate and two rules to follow.
This is called the power rule in most books.
This is how you go:
- multiply by the power and take one away from the power
- Power(x) ( power −1 )
So let’s go through the examples:
Example One
Notice if you have a number there already just times the number by the power. Watch out for x 1/2 and x - 1, it is those negative numbers again!!
f( x) = x 1/2 f’(x) = (1/2 ) x −1/2 = 1 2Ö x
The answer is f’(x) = 1 2Öx
f( x) = x −1 f’(x) = − 1 x −2 = − 1 x2
The answer is f’(x) = −1/ x2
Back to top Back to Product Rule Back to Quotient rule
This is where we have a slightly more complicated looking function than the simple differentiation function. This method I call “The twist” as you use the simple method with a twist at the end.
The function looks more complicated such as 3(x3 + 21 −10x)7 or (x + 4x3)1/2
This is how you go:
- Differentiate as if it is simple, multiply by the power and take one away from power.
- Because it is not simple, here is the twist multiply by the derivative of the bracket.
- Simplify if possible
So let’s go through the examples
Example Two
f (x) = 3(x3 + 21 −10x) 7 f’(x) = (7 ´ 3) (x3 + 21 − 10x) 6 (3x2 − 10)
(multiply by power) (Original bracket) Power- 1(derivative of the bracket)
The answer is f’(x) = 21 (3x2 − 10) (x3 + 21 − 10x) 6
Example Three
f (x) = (x + 4x3) 1/2 f’(x) = 1/2 (x + 4x3) −1/2 (1 + 12x2)
(multiply by power) (Original bracket) Power- 1(derivative of the bracket) (Using Simple differentiation)
The answer is f’(x) = 1/2 (1 + 12x 2) (x + 4x 3) − 1/2
Back to top Back to Product Rule
I find it handy to put brackets around my x-values to help me and especially when differentiation.
This is how you go:
- Put bracket around the expression being logged.
- Use logarithm rules to simplify if possible.
- The answer is a fraction.
- The denominator is the original bracket.
- The numerator is the derivative of the bracket.
So let’s go through the example:
Example Four
f (x) = ln x 3 +5x
f (x) = ln (x 3 +5x)
f’(x) = 3x 2 + 5 Derivative of the bracket (using simple differentiation in this case) x 3 +5x Original Bracket
The answer is f’(x) = 3x 2 + 5 x 3 +5x
Back to top Back to Product Rule Back to Quotient rule
This differentiation always has the original function in the answer, once again I find it handy to put brackets around the power.
This is how you go:
So let’s go through the example:
Example Five
f(x) = e ( −3x)
f’(x) = e ( −3x) . ( − 3 )
f’(x) = e (bracket) .( derivative of the bracket).
The answer is f’(x) = − 3 e ( −3x )
Back to top Back to Product Rule
This is not difficult and in one way is similar to the chain differentiation. We write down dy/dx or f’(x) without saying or thinking that differentiating y or f(x) gives us dy/dx or f’(x) as we are so worried about differentiating the x-values on the other side. This time we have not got “ y = some expression” instead we have the x and y values mixed together.
This is how you go:
- Check to see if you can get the expression into y = ......
- Differentiate the x-values as normal.
- Differentiate the y-values as if they are x-values.
- But the twist is any y-value must be multiplied by dy/dx.
- Move all terms with dy/dx to one side, all other terms on the other side.
- Take dy/dx out as a factor and divide by the bracket.
So let’s go through the example
Example Six
Now we cannot simplify the expression to y = ..... so follow the instructions
2 y (2 - 1) dy/dx + 4 dy/dx + 0 = 6 e6x
(Treat just (multiply by (Treat just (multiply by = (Exponential like x-value) dy/dx) like x-value) dy/dx) Differentiation)
2y dy/dx + 4 dy/dx = 6 e 6x
dy/dx ( 2y + 4 ) = 6 e 6x (Take dy/dx out as a factor)
dy/dx = 6 e 6x ( Divide by the bracket) ( 2y + 4 )
The answer is f’(x) = 6 e 6x ( 2y + 4 )
Back to top
Sixth Way Trigonometrical Differentiation
Ok, I am not going to show how these derivatives are obtained but just give you a list of what the functions become when they are differentiated
Sin (x) differentiates to Cos (x) Cos (x) differentiates to − Sin (x) Tan(x) differentiates to Sec2 (x)
Now we can have a few varieties with trig such as Sin2 (x) and Cos(3x), so what do we do?
To be continued soon
This is a case of following an easy procedure and substituting values into the expression. The Product Rule can involve all the different types of differentiation
This is how you go:
- Put brackets around the two terms (it could be more but not usually).
- Differentiate the terms separately.
- Multiply the first term by the derivative of the second term
- Add the answer above to the second term multiplied by the derivative of the first term.
- This is the answer.
If u = first term then u’ = derivative of first term and v = second term v’ = derivative of second term
The derivative of u v = u v’ + v u’
So let’s go through the example,
This example has a few of the differentiations not all products are as complicated but it is good to see them before a test and not in the test for the first time.
Example Seven
Now you can substitute the values into the expression u v’ + v u’ .
If you set the answer out as above, you may just multply diagonally and add the answers. This is exactly the same as the expression u v’ + v u’ .
d(u v) / dx is the mathematical way of writing the derivative of a product
d(e 3x(x 2 + ln x) 3) / dx = e 3x 3(x 2+ ln x ) 2( 2x +1/x) + (x 2 + ln x) 3 3e 3x
Take out any factors, I will just write f’(x) for d(e 3x(x 2 + ln x) 3) / dx so
We will leave that as the answer but you can always do more simplification if you want.
The answer is f’(x) = 3e 3x (x 2+ ln x ) 2 { 2x + 1/x + (x 2 + ln x) }
Back to top
Again, this is a case of following an easy procedure and substituting values into the expression. The Quotient Rule can involve all the different types of differentiation.
This is how you go:
The derivative of u / v = v u’ - u v’ v2
So let’s go through the example:
Example Eight
Now you can substitute the values into the expression v u’ − u v’ v 2
If you set the answer out as above, starting from top left, you may just multply diagonally and subtract the answers and divide by the square of the bottom. This is exactly the same as the expression v u’ - u v’ v 2
d(u / v) / dx is the mathematical way of writing the derivative of a quotient
d(u / v) / dx = v u’ − u v’
d(ln(3x2+ 4) /x3) / dx = x 3 6x / (3x 2 + 4) − ln (3x 2 + 4) 3x 2
Simplify as much as possible:
To make it easy just take the top by itself and making it one term:
Putting the bottom back in, x6
Cancelling x2, top and bottom
The answer is f’(x) = 3{2x2 − (3x2 + 4) ln (3x2 + 4)}
Back to top
|