What is the dy/dx of y =√ (x^4+3x −1)?

To tackle a problem like this you should use the chain rule. Rewrite this equation as:

(x^4 + 3x -1)^(1/2)

Let u = x^4 + 3x -1

Therefore y = u^(1/2)

dy/dx = dy/du * du/dx

dy/du = (1/2) u ^(-1/2)

du/dx = 4x^3 + 3

substitute u = x^4 + 3x -1 in dy/du

(1/2) (x^4 + 3x -1)^(-1/2) (4x^3 + 3)

This can be rewritten

(4x^3 + 3) / (2√ (x^4 + 3x -1))

I'd square both sides and implicitly derive with the power rule.

y^2 = x^4 + 3x - 1

2y * dy = 4x^3 * dx + 3 * dx

2y * dy = dx * (4x^3 + 3)

dy/dx = (4x^3 + 3) / (2y)

dy/dx = (4x^3 + 3) / (2 * sqrt(x^4 + 3x - 1))

There you go.

Square both sides:

y^2 = x^4 + 3x - 1

Differentiate implicitly:

2yy' = 4x^3 + 3

Divide both sides by 2y:

y' = (4x^3 + 3) / 2y

Now substitute √(x^4 + 3x - 1) back for y:

y' = (4x^3 + 3) / 2√(x^4 + 3x - 1)

y = [ x^4 + 3x - 1 ]^(1/2)

dy/dx = (1/2) [ x^4 + 3x - 1 ]^(-1/2) [ 4x³ + 3]

dy/dx = [ 4x³ + 3 ] / [ 2 ( x^4 + 3x - 1 )^(1/2) ]

y' = dy/dx = (1/2)((x^4+3x-1)^((1/2)-1)) d/dx(x^4+3x-1)

y' = (1/2)((x^4+3x-1)^(-1/2)) (4x^3 + 3)

y' = (1/(2( √ (x^4+3x −1)))) (4x^3 + 3)