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Differentiate following functions.
Solution
At this point there in fact isn't a lot of cause to use the product rule.
We will utilize the product rule. As we add up more functions to our repertoire and as the functions become more complexes the product rule will become more useful and in several cases required.
Note as well that we took the derivative of this function in the previous section and didn't use the product rule at that point. However, we have to get the same result here as we did then.
By converting the radical to a fractional exponent as always, we get.
y = x 2/3 (2 x - x2 )
Now let's take the derivative. Hence we take the derivative of the first function times the second then add up on to that the first function times the derivative of the second function.
y′ = (2/3) x -1/3 (2 x - x2 ) + x 2/3 ( 2 - 2 x )
y′ =(4/3)x(2/3)-(2/3) x(5/3) +2x (2/3) -2x (5/3) =(10/3) x(2/3) -(8/3)x(5/3)
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Indeterminate form : The 0/0 we initially got is called an indeterminate form. It means that we don't actually know what it will be till we do some more work. In the denominator
Now we need to move onto something called function notation. Function notation will be utilized heavily throughout most of remaining section and so it is important to understand i
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