1. ## Very annoying differentials

right ive been given some questions for tomorrow after a 5 month break from university and i have completely lost the technique

1) using product rule find f'(t) of cos t sin t

i manage to get (-sin t)^2 (cos t)^2

2) use quotient rule to find f'(t) of cos t / sin t

once again i manage to get - sin^2 t - cos^2 t / sin^2 t

i cant help but think i should be getting very simple answers?

2. Originally Posted by Revolt
right ive been given some questions for tomorrow after a 5 month break from university and i have completely lost the technique

1) using product rule find f'(t) of cos t sin t

i manage to get (-sin t)^2 (cos t)^2
How do you get that:

d/dt[cos(t)sin(t)] = -sin^2(t) + cos^2(t) = cos(2t)

The product rule is:

d/dx[g(x)h(x)]=g'(x)h(x)+g(x)h'(x)

RonL

3. Originally Posted by Revolt
1) using product rule find f'(t) of cos t sin t

i manage to get (-sin t)^2 (cos t)^2
Another attempt

$f(t)=\cos t\sin t=\frac12\sin2t\implies f'(t)=\cos2t$

4. Originally Posted by CaptainBlack
How do you get that:

d/dt[cos(t)sin(t)] = -sin^2(t) + cos^2(t) = cos(2t)

The product rule is:

d/dx[g(x)h(x)]=g'(x)h(x)+g(x)h'(x)

RonL

i get the bit in bold. is that just another way to write cos (2t) or is there another step involved, my identities are poor :/

have cracked no1 but no2 does not seem to get any simpler :/

5. Originally Posted by Revolt
i get the bit in bold. is that just another way to write cos (2t) or is there another step involved, my identities are poor :/

have cracked no1 but no2 does not seem to get any simpler :/
It is just a bog standard trig identity

RonL

6. Originally Posted by Revolt
2) use quotient rule to find f'(t) of cos t / sin t

once again i manage to get - sin^2 t - cos^2 t / sin^2 t

i cant help but think i should be getting very simple answers?
First do you know the quotient rule?

d/dt[g(t)/h(t)] = [g'(t)h(t) - g(t)h'(t)]/[h(t)]^2

Here g(t)=cos(t), and h(t)=sin(t).

RonL

7. Originally Posted by CaptainBlack
First do you know the quotient rule?

d/dt[g(t)/h(t)] = [g'(t)h(t) - g(t)h'(t)]/[h(t)]^2

Here g(t)=cos(t), and h(t)=sin(t).

RonL
yeah, ive managed to get it down to -1 / sin^2 t

think ill not bother with the rest of the questions lol, will just see the guy tomorrow thanks for the help