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| Derivatives with Power, Product and Quotient Rule |
Newton's Quotient
The derivative of a function is the slope of the tangent. To find the derivative of a function or algebraic expression from first principles, we use Newton's Quotient -- which represents the slope of the line in limit position -- because we limit to zero the horizontal distance between 2 points thus making them one. There are 2 versions of Newton's quotient in common use. One uses "h" to represent the change in x and one uses Dx (delta x). Both of them represent the change in y or f(x) over the change in x.
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Example: Find the derivative of f (x) = x2 – 8x + 15
Solution:
When we do some algebra, we get:
With a simple quadratic function, it's not too bad, but we don't always work with simple functions. After doing a whole bunch of derivatives this way, we begin to see the patterns which evolved into our differentiation rules. Let's apply Newton's Quotient to a general case to develop the power rule for derivatives.
The Power Rule for Derivatives
We let f (x) = x n , where n is a Real number. The derivative then will be:
When we expand ( x + h ) n, we use Pascal's Triangle to determine the coefficients of the terms that are in descending powers of x and ascending powers of "h". We know that the first term
is x n and the second term is n x n – 1 h. The last term of the expansion will be h n. The coefficients of all the terms between will be the entries in the n-th line of Pascal's Triangle -- but we don't care what they are. What matters is the power of h in the terms after the second one.
Here's how we expand and find this limit:
Once we cancel x n with – x n and we factor out "h" from each remaining term, we get:
Now, we cancel the "h" top and bottom then set h = 0. The only remaining non-zero term
is nx n - 1, which tells us that the derivative of y = x n equals n x n – 1 . For example, the derivative of x³ is 3x². If there is a coefficient multiplying x then it is multiplied by "n".
Examples:
a) if f (x) = 7x 5,....... f / (x) = 35 x 4, ............. b) if f (x) = – 9x 8,....... f / (x) = – 72 x 7.
c) if f (x) = 3x - 4,..... f / (x) = – 12 x – 5, ......... d) if f (x) = – ½ x 1 / 2,.. f / (x) = – ¼ x – 1 / 2.
warning: lots of students mess up with negative exponents. Make sure to reduce the power by one. If n is a fraction -- say ¾ -- when subtracting 1 -- think of it as 4 / 4 -- so the difference will be – ¼ . In example d) above, we subtract 2/2 from ½ to get – ½.
| The Power Rule for Derivatives If a and n are Real numbers and f (x) = a x n then f '(x) = a n x n – 1 . |
Since any constant c is the same as cx 0, the derivative = 0.
Since cx is the same as c x¹, the derivative = c.
The Product Rule
If we have to differentiate f (x) = 7x 5 ( x² – 3) -- a product of 2 functions -- we can easily multiply to get f (x) = 7x 7 – 21 x 5 , and then use the power rule to find f '(x) = 49 x 6 – 105 x 4 . However, when the functions in the product are more complicated, we use the Product Rule for Derivatives that says:
To find the derivative of a product, take derivative of the first function multiplied by the last plus derivative of the last function multiplied by the first. We use "u" and "v" to represent the two functions. Symbolically, this rule is ( u v )' = u'v + v'u.
In the example above, u = 7x 5 and v = ( x² – 3). So, u' = 35x 4 , and v' = 2x
Using the Product Rule we'll get:
( u v )' = 35x 4(x² – 3) + 2x (7x 5) =
35 x 6 – 105 x 4 + 14 x 6 or 49 x 6 – 105 x 4
Here's an example with more complicated functions for u and v.
Find the derivative of f (x) = (3x³ + 5x) ( 2x² – 3x )
Here, u = (3x³ + 5x) , and v = ( 2x² – 3x ), so u' = ( 9x ² + 5 ), and v' = ( 4x – 3 )
u'v + v'u = ( 9x ² + 5 )( 2x² – 3x ) + ( 4x – 3 )(3x³ + 5x)
Once we multiply out the brackets and collect like terms we get:
f '(x) = 30 x 4 – 36 x³ + 30 x² – 30 x
| The Product Rule for Derivatives If u = f ( x ), and v = g ( x ), then the derivative of the product is: ( uv )' = u'v + v'u which is the derivative of u times v unchanged plus the derivative of v times u unchanged. |
BEWARE!! Many text books today present this rule in the opposite order. They list the derivative of a product as uv' + vu'. Since addition is commutative, it doesn't matter. However, the Quotient Rule involves subtraction of these terms and so order does matter. If we learn the correct order for the product rule, it's much easier to remember the quotient rule.
If your textbook gives you uv' + vu', ignore it and learn the right order for this rule.
The Quotient Rule
If we have to differentiate a quotient of 2 functions, we use the Quotient Rule for Derivatives that says:
To find the derivative of a quotient, take derivative of the numerator times the denominator, minus the derivative of the denominator times the numerator, all divided by the square of the denominator. We use "u" and "v" to represent the two functions. Symbolically, this rule is:
It is best to do the v² part in the denominator before anything else because we tend to forget it once we start differentiating the top and bottom functions.
Example: Find the derivative of 
. Write the equation of the tangent to this function at the point where x = 1.
Solution: Here, u = 3x 4 + 2x , and v = 5x² – x , so u' = 12x³ + 2 , and v' = 10x – 1.
Applying the Quotient Rule for Derivatives, we get:
Once we multiply out the brackets and collect like terms we get:
Now, when x = 1, y = 5/4 -- that's the point of contact. Since the derivative is the slope of the tangent, and the numerical value of the derivative at x = 1 is 11/16, we can write the equation of the tangent to this curve at x = 1 as:
When we do the algebra, we get 
| The Quotient Rule for Derivatives If u = f ( x ), and v = g ( x ), then the derivative of the quotient is:
which is the derivative of u times v unchanged minus the derivative of v times u unchanged, all divided by v ². |
Don't be fooled into thinking that 
is a quotient! The numerator is a constant! Not a function, therefore rewrite the function as –7x – 5 , then use the power rule.
A fraction like 
, so it too is not a quotient. It's just a linear function, so we can use power rule here too.
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Practice Exercises
1) Use the power rule to find the derivative:
| a) f (x) = 5 – 3x4 + 9x | b) f (x) = 2x – 6 – 9x³ | c) f (x) = 10x² + 9x 5. |
| d) f (t) = 12 – 3t² – 5t – 7 | e) f (r) = 21r – 1/3 – 10r 4/5 . | f) f (x) = 4 – 8x 9/4 + 12x – ½. |
2) Use the product rule to find the derivative:
| a) f (x) = (x² – 7)(2x³ + 3) | b) f (x) = (2x² – 4x + 1)(6x – 5) | c) f (x) = (4x – 1 + 9)(x 5 + 1). |
3) Use the quotient rule to find the derivative:
| a) |
b)
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c)
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4) Write the equation of the tangent to the given curve at the given point P:
a) y = 2x³ + 4x² – 5x – 3, ........ P( – 1, 4 ) .
b) y = (3x – 1) / x² ........ ........ P ( 1, 2 ).
c) y = 5 / (1 + x²) ........ ........ P ( – 2, 1 ).
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Solutions
1) Use the power rule to find the derivative:
| a) f (x) = 5 – 3x4 + 9x f '(x) = – 12x³ + 9 |
b) f (x) = 2x – 6 – 9x³ f '(x) = – 12x – 7 – 27x² |
c) f (x) = 10x² + 9x 5. f '(x) = 20x + 45x 4. |
| d) f (t) = 12 – 3t² – 5t – 7 f '(t) = – 6t + 35t – 8 . |
e) f (r) = 21r – 1/ 3 – 10r 4/ 5 f '(r) = – 7r – 4/ 3 – 8r –1/ 5 . |
f) f (x) = 4 – 8x 9/4 + 12x – ½. f '(x) = – 18x 5/ 4 – 6x – 3/ 2 . |
2) Use the product rule to find the derivative:
| a) f (x) = (x² – 7)(2x³ + 3) f '(x) = 2x(2x³ + 3) + 6x²(x² – 7) f '(x) = 10x 4 – 42x² + 6x |
b) f (x) = (2x² – 4x + 1)(6x – 5) f '(x) = (4x – 4)(6x – 5) + 6(2x² – 4x + 1) f '(x) = 36x² – 68x + 26 |
| c) f (x) = (4x – 1 + 9)(x 5 + 1). f '(x) = – 4/x² (x 5 + 1) + 5x 4 (4x – 1 + 9) f '(x) = 45x 4 + 16x³ – 4/x² |
3) Use the quotient rule to find the derivative:
a) 
b) 
c) 
4) Write the equation of the tangent to the given curve at the given point P:
a) y = 2x³ + 4x² – 5x – 3, ........ P( – 1, 4 ) .
y' = 6x² + 8x – 5 = – 7 when x = – 1. Equation of tangent: y = – 7x – 3
b) y = (3x – 1) / x² ........ ........ P ( 1, 2 ).
. Equation of tangent: y = – x + 3
c) y = 5 / (1 + x²) ........ ........ P ( – 2, 1 ).
. Equation of tangent: 4x – 5y + 13 = 0
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