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- Thread starter ocgirl83
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- Feb 4, 2004
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I'm going to guess that you mean the following:
. . . . .\(\displaystyle \Large{y\mbox{ }=\mbox{ }\frac{x^3\mbox{ }+\mbox{ }x\mbox{ }+\mbox{ }2}{x^2\mbox{ }+\mbox{ }3}}\)
If so, then just apply the Quotient Rule:
. . . . .\(\displaystyle \Large{h(x)\mbox{ }=\mbox{ }\frac{f(x)}{g(x)}}\)
. . . . .\(\displaystyle \Large{h'(x)\mbox{ }=\mbox{ }\frac{f'(x)g(x)\mbox{ }-\mbox{ }f(x)g'(x)}{g^2(x)}}\)
...where, in your case:
. . . . .\(\displaystyle \Large{f(x)\mbox{ }=\mbox{ }x^3\mbox{ }+\mbox{ }x\mbox{ }+\mbox{ }2}\)
. . . . .\(\displaystyle \Large{g(x)\mbox{ }=\mbox{ }x^2\mbox{ }+\mbox{ }3}\)
So differentiate the numerator and denominator separately, and then just plug into the formula. That's all there is to it!
Eliz.
. . . . .\(\displaystyle \Large{y\mbox{ }=\mbox{ }\frac{x^3\mbox{ }+\mbox{ }x\mbox{ }+\mbox{ }2}{x^2\mbox{ }+\mbox{ }3}}\)
If so, then just apply the Quotient Rule:
. . . . .\(\displaystyle \Large{h(x)\mbox{ }=\mbox{ }\frac{f(x)}{g(x)}}\)
. . . . .\(\displaystyle \Large{h'(x)\mbox{ }=\mbox{ }\frac{f'(x)g(x)\mbox{ }-\mbox{ }f(x)g'(x)}{g^2(x)}}\)
...where, in your case:
. . . . .\(\displaystyle \Large{f(x)\mbox{ }=\mbox{ }x^3\mbox{ }+\mbox{ }x\mbox{ }+\mbox{ }2}\)
. . . . .\(\displaystyle \Large{g(x)\mbox{ }=\mbox{ }x^2\mbox{ }+\mbox{ }3}\)
So differentiate the numerator and denominator separately, and then just plug into the formula. That's all there is to it!
Eliz.