Question

I'm trying to find the derivative of this function:

http://www.wolframalpha.com/input/?i=derivative+of+-t%2F%289-t%5E2%29%5E1%2F2

Answer 1

Let me know if you can't see it. I understand how they derivated the function I'm just unsure how they got the 9 for the numerator

Answer 2

product rule

Answer 3

and quotient rule

Answer 4

\[-t/(9-t^2)^{1/2}=-t\ (9-t^2)^{-1/2}\]

Answer 5

denominator is
\[g(t)=\sqrt{9-t^2}\] numerator is
\[f(t) =-t\] and so using the quotient rule your new numerator should be
\[gf'-fg'\]

Answer 6

Let t = 3sinx

Answer 7

I think that -t should be a -1

Answer 8

\[D[-t\ (9-t^2)^{-1/2}]=-t'\ (9-t^2)^{-1/2}-t\ (9-t^2)' ^{-1/2}\]

\[-(9-t^2)^{-1/2}+\frac12t\ (9-t^2)^{-3/2}*(9-t^2)'\]

\[-(9-t^2)^{-1/2}+\frac12t\ (9-t^2)^{-3/2}*(-2t)\]

\[-(9-t^2)^{-1/2}-t^2\ (9-t^2)^{-3/2}\]

\[(-1-t^2\ (9-t^2)^{-1})\ (9-t^2)^{-1/2}\]

\[\frac{-1-t^2\ (9-t^2)^{-1}}{(9-t^2)^{1/2}}\]

\[\frac{-1}{(9-t^2)^{1/2}}+\frac{-t^2\ (9-t^2)^{-1}}{(9-t^2)^{1/2}}\]

\[\frac{-1}{(9-t^2)^{1/2}}+\frac{-t^2}{(9-t^2)^{3/2}}\]

\[\frac{-(9-t^2)}{(9-t^2)^{3/2}}+\frac{-t^2}{(9-t^2)^{3/2}}\]

\[\frac{-9+t^2-t^2}{(9-t^2)^{3/2}}\]

Answer 9

oh okay now it makes sense to me thank you.

Answer 10

I just type my short version:

Answer 11

= - [( √9-t² )/ (√9-t² ) ]/ ( 9 - t²)
= - ( 9 - t² ) + t² / ( 9 - t² ) (√9+t² )
= -9/ √(9+t²)³

Answer 12

thats even better. Thanks.

Answer 13

Because my mind is short short :P