Derivatives: z = (9 - 7t) / (10 + t)

A

Angela123

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Oct 9, 2008
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z=(9-7t)/(10+t)

I did: (9-7t)+7/(10+t)^2 and ended up with [(9-7t)+7]/[2(10+t)]. Where did I go wrong?
 
The Quotient Rule for f(x) = g(x)/h(x) is:

. . . . .\(\displaystyle f'(x)\, =\, \frac{g'(x) h(x)\, -\, g(x)h'(x)}{h^2(x)}\)

Letting g(t) = 9 - 7t and h(t) = 10 + t, I don't see how you got your numerator in your first step...?
 
Ok I tried doing it again and got: (10+t)-(9-7t)/(2(10+t)) What's wrong with it this time?
 
Angela123 said:
Ok I tried doing it again and got: (10+t)-(9-7t)/(2(10+t)) What's wrong with it this time?
Click to expand...
I will assume that you forgot the grouping symbols, and that you actually mean (10 + t) to be included within the numerator.

But the first term in the numerator is supposed to get g'(t)h(t). I see h(t), but no evidence of g'(t).

The second term in the numerator is supposed to be g(t)h'(t). I see g(t), but I'm not sure that you've accounted for h'(t).

The denominator is supposed to be h[sup:14h3e3q5]2[/sup:14h3e3q5](t), but you appear instead to have used 2h(t)...?

Try doing this step by step. What is g(t)? What is g'(t)? What is h(t)? What is h'(t)? What is [h(t)][sup:14h3e3q5]2[/sup:14h3e3q5]? Write these all out.

Once you have expressions for each of these bits, plug them into the formula provided by the Quotient Formula. See what you get.
 
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