How do you find d/dt given #y=t^3 - 3t# and #x=t^2#?

1 Answer
Oct 5, 2016

It depends upon what you intended by #d/(dt)#

See possible interpretations below:

Explanation:

Option 1: You wanted #color(black)((dcolor(red)y)/(dt))#
#(dy)/(dt)=3t^2-3# (using the exponent rule)

Option 2: You wanted #color(black)((dcolor(red)x)/(dt))#
#(dx)/(dt)=2t# (using the exponent rule)

Option 3: You wanted #color(black)((dy)/(dcolor(red)x)#
...definitely not what was asked for, but a possibly intended requirement.
If #y=t^3-3t# and #x=t^2#
then #t=x^(1/2)# and #t^3=x^(3/2)#
so
#y=x^(3/2)-3x^(1/2)#
and
#(dy)/(dx)= 3/2x^(1/2) -3/2x^(-1/2) = (3sqrt(x))/2-3/(2sqrt(x))#