What is the d/(dy)(sqrt(x^2+4xy+y^6)-8) ?

The d/(dy)(sqrt(x^2+4xy+y^6)-8) is (2x+3y^5)/(sqrt(x^2+4xy+y^6))

What is the first d/(dy)(sqrt(x^2+4xy+y^6)-8) ?

The first d/(dy)(sqrt(x^2+4xy+y^6)-8) is (2x+3y^5)/(sqrt(x^2+4xy+y^6))

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\frac{d}{d y} (x^{2} + 4 x y + y^{6} - 8)

\frac{2 x + 3 y ^{5}}{x ^{2} + 4 x y + y ^{6}}

Solution steps

\frac{d}{d y} (x^{2} + 4 x y + y^{6} - 8)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d y} (x^{2} + 4 x y + y^{6}) - \frac{d}{d y} (8)

\frac{d}{d y} (x^{2} + 4 x y + y^{6}) = \frac{3 y ^{5} + 2 x}{x ^{2} + 4 x y + y ^{6}}

\frac{d}{d y} (8) = 0

= \frac{3 y ^{5} + 2 x}{x ^{2} + 4 x y + y ^{6}} - 0

Simplify

= \frac{2 x + 3 y ^{5}}{x ^{2} + 4 x y + y ^{6}}

\frac{d}{d x} (x^{3} + 6 x^{2} + 6 x)

f (x) = 2 - 5 x^{2}

\frac{3 s}{( s + 2 ) ( s - 4 )}

x = 5 y^{2}, x = 2 + 3 y^{2}

\frac{d ^{2}}{d x ^{2}} (\frac{2 t - 1}{t ^{4}})

What is the d/(dy)(sqrt(x^2+4xy+y^6)-8) ?
The d/(dy)(sqrt(x^2+4xy+y^6)-8) is (2x+3y^5)/(sqrt(x^2+4xy+y^6))
What is the first d/(dy)(sqrt(x^2+4xy+y^6)-8) ?
The first d/(dy)(sqrt(x^2+4xy+y^6)-8) is (2x+3y^5)/(sqrt(x^2+4xy+y^6))