The Product Rule f ' ( x ) = v ' ( x ) · u ( x ) + u ' ( x ) · v ( x ) is probably the rule we will use the most in conjunction with the other types of derivatives. For example, finding u ' ( x ) may involve one rule while finding v ' ( x ) may involve another rule. Example: y = 8 x 2 · ln ( 4 x 2 – 3 x + 2 ) Solution: We first determine that the main rule to follow is the product rule. Therefore u ( x ) = 8 x 2 and v ( x ) = ln ( 4 x 2 – 3 x + 2 ) We determine to use the power rule to find u ' and Dx Dx [ ln | g ( x ) | ] to find v '. u ' ( x ) = 16 x g ( x ) = 4 x 2 – 3 x + 2, then g ' ( x ) = 8 x – 3

Now substitute these answers into the product rule f ' ( x ) = v ' ( x )  u ( x ) + u ' ( x )  v ( x ). Let us work a problem that looks similar but uses a different set of rules. Solution: We first determine that the main rule to follow is the product rule. Therefore and v ( x ) = ln ( 4 x 2 – 3 x + 2 ) Dx Dx [ | g ( x ) | ] to find v '. Then use the derivative of a g (x) (x) which is D x [ a g (x) (x) ] = ( ln a) a) [ g ' ( x ) ] a g (x) (x) to find u ' ( x ). and a = 8 g ( x ) = x 2 g ' ( x ) = 2 x g ( x ) = 4 x 2 – 3 x + 2, then g ' ( x ) = 8 x – 3

You can not multiply these numbers together. Now substitute these answers into the product rule f ' ( x ) = v ' ( x )  u ( x ) + u ' ( x )  v ( x ).

In mathematics f ( x ), g ( x ), etc. are just names. If it becomes to confusing using the same name in a problem you may want to change one of the names to h ( x ) or some other functional name. I will do that in the next example. Solution: We first determine that the main rule to follow is the product rule. Therefore and Use the derivative of e g (x):(x): D x [ e g (x) (x) ] = g ' ( x ) · e g (x)(x) and Use the derivative of log a | h ( x ) | which is g ( x ) = 4 x 2 – 5, then g ' ( x ) = 8 x a = 3 h ( x ) = 2x 2x 2 – 3 x + 1 h ' ( x ) = 4 x – 3

Now substitute these answers into the product rule f ' ( x ) = v ' ( x )  u ( x ) + u ' ( x )  v ( x ).

Use the derivative of ln | g ( x ) | which is To find g ' ( x ) use the chain rule alternative form but to reduce some of the confusion with all of the g ( x )’s and g ' ( x )’s let us change the notation to = h ' ( x ) · f ' [ h ( x ) ]. Let h ( x ) = 6x 6x 2 – 5 x + 7 h ' ( x ) = 12 x – 5 ( 6x 2 – 5 x + 7 ) Therefore Substituting answers into the derivative:

Rules of algebra when canceling like terms: take the smallest exponent from the largest and write the answer where the largest exponent existed. The exponent of one is not written because it is understood. Compare this new problem with the previous problem. They may look similar but there are subtle differences. Do not try to generalize or take any false short cuts. Make certain you understand the algebra rule of exponents and arithmetic. Determine to use the derivative of ln | g ( x ) | is To find g ' ( x ) use the chain rule alternative form. = h ' ( x ) · f ' [ h ( x ) ]. Let h ( x ) = 7x 7x x – 4 h ' ( x ) = 14 x + 3

Therefore

I suggest that you try to solve these problems in small steps until you become very familiar with each type. Try to refrain from trying to take short cuts or immediately trying to write the final answers. It may take a little longer but probably you will be more successful. Speed only comes from many hours of practice. u ( x ) = 9 x + 2 v ( x ) = ( 2 x 2 – 7 x + 3 ) 2 u ' ( x ) = 9 v ' ( x ) = 2 ( 4 x – 7 )( 2 x 2 – 7 x + 3 ) Now substitute these answers into the product rule f ' ( x ) = v ' ( x )  u ( x ) + u ' ( x ) ) v ( x ). y ' = 2 ( 4 x – 7 )( 2 x 2 – 7 x + 3 )( 9 x + 2 ) + 9 ( 2 x 2 – 7 x + 3 ) 2 y ' = ( 2 x 2 – 7 x + 3 ) [ 2 ( 4 x – 7 )( 9 x + 2 ) + 9 ( 2 x 2 – 7 x + 3 ) ] y ' = ( 2 x 2 – 7 x + 3 ) [ ( 8 x – 14 )( 9 x + 2 ) + 9 ( 2 x 2 – 7 x + 3 ) ] y ' = ( 2 x 2 – 7 x + 3 ) [ 72 x x – 126 x – x 2 – 63 x + 27 ] y ' = ( 2 x 2 – 7 x + 3 ) ( 90 x 2 – 173 x – 1 )