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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Originally Posted by Truthbetold Fin dy/dx $\displaystyle y=\frac{200}{1+\varepsilon^{5-t}}$ That's e^(5-t) I'm working to fix the Latex $\displaystyle \frac{\,dy}{\,dx} =0$... But: $\displaystyle \frac{\,dy}{\,dt} = \frac{(1+e^{5-t})\frac{d}{dt}(200)-200\frac{d}{dt}(1+e^{5-t})}{(1+e^{5-t})^2}$ $\displaystyle =\frac{(0)-200(-e^{5-t})}{(1+e^{5-t})^2}$ $\displaystyle =\frac{200e^{5-t}}{(1+e^{5-t})^2}$
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