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.
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)
2 answers
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
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