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
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
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