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

The d/(dy)((x)(x^3+y^4)) is 4xy^3

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

The first d/(dy)((x)(x^3+y^4)) is 4xy^3

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

4 x y^{3}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

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

= x (0 + 4 y^{3})

Simplify

= 4 x y^{3}

\frac{( - x + 11 ) ( - x + 5 )}{2}

\frac{( t ^{2} + 1 ) ( 1 - t )}{t ^{3} + 1}

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

ln (a + e^{(x + y + z)^{2}})

(2 - 6) a + (3 - 6) b

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