Given y= f(t) = 2/t, find the derivative of y with respect to t.
the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\) always
unless you are supposed to do this by hand
If \[ y=t^n\] then \[ y'= n t^{n-1} \]
Apply the above for n=-1;
umm. (2/t)^-1? I think I am supposed to do this using the limit formula though.
that is what i suspected
|dw:1475803407611:dw|
this correct?
\[\frac{\frac{1}{t+h}-\frac{1}{t}}{h}\]
then a bunch of algebra, subtract first up top
yes that is the formula. @satellite73 i know the answer to this question so what you said earlier about 1/x makes sense. I just don't know how to do this, and the teacher used the formula to solve this. and ok, 2t-2(t+h) ---------- t(t+h) ?
sure looks good
numerator is just \(-2h\) when you distribute and combine like terms
so you are looking at \[\frac{-2h}{t(t+h)}\] now divide by \(h\)
-2h/th(t+h) ?
cancel the h top and bottom
gives \[\frac{-2}{t(t+h)}\] then put \(h=0\) to get the answer you know it has to be
ohhh. oh i forgot about substituting. ok. -2/t(t+0) = -2/t^2 thank you so much.
yw
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