Question 328484
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H(t)\ =\ \frac{5000}{t^2\ +\ 10}]


Use the Quotient Rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H'(t)\ =\ \frac{\left(t^2\ +\ 10\right)\ -\ 10000t}{\left(t^2\ +\ 10\right)^2}]


You want


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H'(3)\ =\ \frac{\left((3)^2\ +\ 10\right)\ -\ 10000(3)}{\left((3)^2\ +\ 10\right)^2}]


You can do your own arithmetic.


You can also use the chain rule on this one:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{5000}{t^2\ +\ 10}]


Let *[tex \Large u\ =\ t^2\ +\ 10]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{5000}{u}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{du}\ =\ -\frac{5000}{u^2}\ =\ -\frac{5000}{\left(t^2\ +\ 10\right)^2]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{du}{dt}\ =\ 2t]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \left(\frac{dy}{du}\right)\left(\frac{du}{dt}\right)\ =\ -\frac{10,000t}{\left(t^2\ +\ 10\right)^2]


And again, you are seeking the value of


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H'(3)=\ -\frac{10,000(3)}{\left((3)^2\ +\ 10\right)^2]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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