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

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

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

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

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

4 x^{2} y

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d y} (2 y^{2} x) = 4 x y

\frac{d}{d y} (3) = 0

= x (4 x y - 0)

Simplify

x (4 x y - 0) : 4 x^{2} y

= 4 x^{2} y

\frac{1}{2 ( 3 y ^{5} - 1 )}

\frac{( x + 1 )}{2 ( x - 1 )}

\frac{2 ( 1 + x ) 1 + x}{3}

(x + 1) (x + 4 = x^{2} + 14)

\frac{1}{x ( 1 + x ) ^{2}}

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