I want to derivate this function :
$$f(t) = \frac{3}{\sin(t)}$$
I know that the derivative of $\frac{u(x)}{v(x)}$is$\frac{u'v-uv'}{v^{2}}$ in general and that in this fraction :
$$u'(t) = 0$$
$$v'(t) = \cos(t)$$
So I do :
$$\frac{0\times sin(t)-0\times cos(t)}{\sin(t)^{2}} = 0$$
But Wolfram Alpha gives me another result :
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Jasser
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Pop Flamingo
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3your $u$ term is $3$ not $0$. – James Sep 14 '14 at 14:17
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@James Oh thank you ! – Pop Flamingo Sep 14 '14 at 14:18
2 Answers
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There's an error somewhere in your calculations
$$ \frac{d}{dt} \left(\frac {3}{\sin(t)} \right) = \frac {\frac{d}{dt}3 \sin(t)-3\frac{d}{dt}\sin(t)}{\sin^2(t)} = \frac {0 \sin(t)-3\cos(t)}{\sin^2(t)} = -\frac {3cos(t)}{\sin^2(t)} $$
patatahooligan
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Did you made an error ? I wrote $\frac{3\cos(t)}{\sin^{2}(t)}$ and I thought it was $\frac{-3\cos(t)}{\sin^{2}(t)}$ – Pop Flamingo Sep 14 '14 at 14:22
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Because in the formula of the lesson it's a subtraction, not an addition... – Pop Flamingo Sep 14 '14 at 14:23
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Yours too. $$\frac {\frac{d}{dt}(3) \sin(t)\color{red}{-}3\frac{d}{dt}\sin(t)}{\sin^2(t)}$$ – UserX Sep 14 '14 at 14:23
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Yeah, it was supposed to be $-$. Thanks for bringing it to my attention, both of you. – patatahooligan Sep 14 '14 at 14:25