Question

Find the x' for x(t) defined implicity by x^3+t^3x+t^4-8=0.Then evaluate x' at the point (-2,2)

Answer 1

x^3+t^3x+t^4-8=0

d/dt [ x^3+t^3x+t^4-8] = d/dt[ 0 ]

d/dt [ x^3 ] + d/dt [ t^3x ] + d/dt [ t^4 ] -d/dt [ 8 ] = 0

3x^2 * x ' + d/dt [ t^3x ] + 4t^3 = 0

3x^2 * x ' + 3t^2*x + t^3 * x ' + 4t^3 = 0

Now solve for x '

Answer 2

okay give me a sec

Answer 3

i still dont understand a little lol

Answer 4

3x^2 * x ' + 3t^2*x + t^3 * x ' + 4t^3 = 0

3x^2 * x ' + t^3 * x ' = -4t^3 - 3t^2*x

x ' ( 3x^2 + t^3 ) = -4t^3 - 3t^2*x

I'll let you finish