Table of Derivatives

legacyofpiracy

Junior Member
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Oct 20, 2005
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I just got stuck on another problem..although i have gotten a few blanks filled in . The problem was to fill in a table by using the product and quotient rules : h(x)=f(x)*g(x) and j(x)=f(x)/g(x).


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I was already able to calulate the numbers that are highlighted but perhaps someone could just double check. I still, however, can't figgure out how to caluclate the two numbers in the middle collum (x=1) with the given information. Any thoughts?
 
h’(x)=f’(x)g(x)+f(x)g’(x) & j’(x)=[f’(x)g(x)−f(x)g’(x)]/[g(x)]<SUP>2</SUP>.
Now use the values from the table to finish it off.
 
Using those two formulas is how i calculated the highlighted numbers, however I can't figgure out how to determine those remaining two. IN order to submit them into the equation i would need to have at least one of the g'(x) or f'(x) values wouldn't i?
 
You have h(x) = f(x)g(x), so h'(x) = f'(x)g(x) + f(x)g'(x). To find f'(1), plug in "3" for f(1), "3" for g(1), and "-6" for h'(1). This gives you -6 = 3f'(x) + 3g'(x).

You also have j(x) = f(x)/g(x), so j'(x) = [f'(x)g(x) - f(x)g'(x)] / [g(x)]<sup>2</sup>. Plug in "2" for j'(1), "3" for f(1), and "3" for g(1). This gives you 2 = [3f'(x) - 3g'(x)] / [3]<sup>2</sup>.

Considering the two together, you get two equations in two unknowns, something you can solve.

. . . . .-6 = 3f' + 3g'
. . . . .2 = [3f' - 3g'] / 9

. . . . .-6 = 3f' + 3g'
. . . . .18 = 3f' - 3g'

Adding down, you get:

. . . . .12 = 6f'
. . . . .2 = f'

So f'(1) = 2. Then:

. . . . .-6 = 3(2) + 3g'
. . . . .-6 = 6 + 3g'
. . . . .-12 = 3g'
. . . . .-4 = g'

So g'(1) = -4.

Hope that helps a bit.

Eliz.
 
h’(1)=f’(1)g(1)+f(1)g’(1)
j’(1)=[f’(1)g(1)−f(1)g’(1)]/[g(1)]<SUP>2</SUP>
Can you solve this system?
 
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