Posted by **Christine ** on Sunday, February 13, 2011 at 8:14am.

Can anyone help with this...I need to find the derivative of the functions below. If possible please show working so I can try and understand?

f(t) =3-t^4 and g(t)=sin(4t)

Then using the Quotient Rule differentiate the function

k(t) 3-t^4/ sin(4t) (0<t< pie)

- Maths - Calculus -
**bobpursley**, Sunday, February 13, 2011 at 8:25am
f'=d3/dt-d/dt (t^4=0-4t^3

g'=d/dt (sin(4t))= cos4t * d4t/dt=4cos4t

ARRRRGGG. Quotent rule in ASCII

That is too much algebra for me to do now, maybe later.

- Maths - Calculus -
**Reiny**, Sunday, February 13, 2011 at 9:09am
2nd question:

first line derivative

= [ -4t^3(sin(4t)) - 4cos(4t)(3-t^4) ] / (sin(4t))^2

after that there are several things you could do, I don't know what kind of answer your course is expecting.

you could split it up into two fractions ...

= (-4t^3)/sin(4t) - 4(3-t^4)cot(4t) / sin(4t)

as one possibility

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