Find dy/dx
6. y^3=x^2+1
Im having trouble understanding dy/dx problems.
3 answers
I feel your pain, buddy, but I don't know calculus
Us ethe chain rule.
3y^2 y' = 2x
y' = 2x/3y^2
Or, if you want it explicitly,
y = (x^2+1)^(1/3)
y' = 1/3 (x^2+1)^(-2/3) * 2x
you can see that the two are the same.
3y^2 y' = 2x
y' = 2x/3y^2
Or, if you want it explicitly,
y = (x^2+1)^(1/3)
y' = 1/3 (x^2+1)^(-2/3) * 2x
you can see that the two are the same.
take d/dx of both sides
3 y^2 dy/dx = 2 x + 0 ( because dx/dx is 1)
dy/dx = (2/3) x/y^2
or
dy/dx = (2/3) x/ (x^2+1)^(2/3)
3 y^2 dy/dx = 2 x + 0 ( because dx/dx is 1)
dy/dx = (2/3) x/y^2
or
dy/dx = (2/3) x/ (x^2+1)^(2/3)
3 answers