Fin dy/dx

$\displaystyle y=\frac{200}{1+\varepsilon^{5-t}}$

That's e^(5-t) I'm working to fix the Latex

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- Dec 10th 2007, 08:30 PMTruthbetoldDerivative with e
Fin dy/dx

$\displaystyle y=\frac{200}{1+\varepsilon^{5-t}}$

That's e^(5-t) I'm working to fix the Latex - Dec 10th 2007, 08:53 PMDivideBy0
$\displaystyle \frac{\,dy}{\,dx} =0$...

But:

$\displaystyle \frac{\,dy}{\,dt} = \frac{(1+e^{5-t})\frac{d}{dt}(200)-200\frac{d}{dt}(1+e^{5-t})}{(1+e^{5-t})^2}$

$\displaystyle =\frac{(0)-200(-e^{5-t})}{(1+e^{5-t})^2}$

$\displaystyle =\frac{200e^{5-t}}{(1+e^{5-t})^2}$