What is the d/(dy)(y^3+kxy^4-2x) ?

The d/(dy)(y^3+kxy^4-2x) is 4ky^3x+3y^2

What is the first d/(dy)(y^3+kxy^4-2x) ?

The first d/(dy)(y^3+kxy^4-2x) is 4ky^3x+3y^2

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\frac{d}{d y} (y^{3} + k x y^{4} - 2 x)

4 k y^{3} x + 3 y^{2}

Solution steps

\frac{d}{d y} (y^{3} + k x y^{4} - 2 x)

Treat

k, x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d y} (y^{3}) + \frac{d}{d y} (k x y^{4}) - \frac{d}{d y} (2 x)

\frac{d}{d y} (y^{3}) = 3 y^{2}

\frac{d}{d y} (k x y^{4}) = 4 x k y^{3}

\frac{d}{d y} (2 x) = 0

= 3 y^{2} + 4 x k y^{3} - 0

Simplify

= 4 k y^{3} x + 3 y^{2}

\int \frac{1}{x ^{2}} sin (\frac{1}{x}) cos (\frac{1}{x}) d x

\int_{0}^{3} x f (x^{2}) d x

\int_{0}^{2} 4 x^{2} - 2 x^{3} d x

\frac{d}{d x} (x (x - 3)^{2})

y = x^{2}, y = - x^{2} + 4 x

What is the d/(dy)(y^3+kxy^4-2x) ?
The d/(dy)(y^3+kxy^4-2x) is 4ky^3x+3y^2
What is the first d/(dy)(y^3+kxy^4-2x) ?
The first d/(dy)(y^3+kxy^4-2x) is 4ky^3x+3y^2