calc derivative help... please?

Question

derivative of

f(t)=(2t)/(4+t^2)

y= (t^3 + t)/ (t^4-2)

y=( x^(1/2)-1)/ (x^(1/2)+1)

Answer 1

for the first one, use the quotient rule:
numerator times the derivative of denominator minus denominator times the derivative of numerator.. all over the denominator squared.

f'(t) = [ 2t (2t) - (4+t^2)(2) ] / [4+t^2]^2
=( 4t^2 - 8 - 2t^2 ) / [4+t^2]^2
=( 2t^2 - 8 ) / [4+t^2]^2
that should be it, if you think it's not simplified further.. i'm sure you can do it on your own. =]

you use the quotient rule on the rest.. just make sure that when it's to the 1/2 power, it's the same as the squareroot of it..

Answer 2

f(t)=(2t)/(4+t^2)
f '(t) = [2(4+t^2)-2t(2t)]/(4+t^2)^2
= 2(4- t^2)/(4+t^2)^2

y= (t^3 + t)/ (t^4-2)
y' =[(t^4-2)(3t^2+1) -(t^3+t)(4t^3)]/(t^4-2)^2
y' = I'll let you simplify

y=( x^(1/2)-1)/ (x^(1/2)+1)
y' = [(x^.5 +1)(1/(2x^.5)) - (x^.5-1)(1/(2x^.5)]/(x^.5+1)^2
y' = 1/[(x^.5)(x^.5 +1)^2]