What is (\partial)/(\partial x)(sqrt(x^2+8xy+y^2)) ?

The answer to (\partial)/(\partial x)(sqrt(x^2+8xy+y^2)) is (x+4y)/(sqrt(x^2+8xy+y^2))

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(\partial)/(\partial x)(sqrt(x^2+8xy+y^2))

\frac{\partial}{\partial x} (x^{2} + 8 x y + y^{2})

\frac{x + 4 y}{x ^{2} + 8 x y + y ^{2}}

Solution steps

\frac{\partial}{\partial x} (x^{2} + 8 x y + y^{2})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 x ^{2} + 8 x y + y ^{2}} \frac{\partial}{\partial x} (x^{2} + 8 x y + y^{2})

= \frac{1}{2 x ^{2} + 8 x y + y ^{2}} \frac{\partial}{\partial x} (x^{2} + 8 x y + y^{2})

\frac{\partial}{\partial x} (x^{2} + 8 x y + y^{2}) = 2 x + 8 y

= \frac{1}{2 x ^{2} + 8 x y + y ^{2}} (2 x + 8 y)

Simplify

\frac{1}{2 x ^{2} + 8 x y + y ^{2}} (2 x + 8 y) : \frac{x + 4 y}{x ^{2} + 8 x y + y ^{2}}

= \frac{x + 4 y}{x ^{2} + 8 x y + y ^{2}}

x^{s i n (x)}

f (x) = x^{3} + x^{2} - x - 1, at x = 0

\frac{d y}{d t} - 4 y = 10 sin (2 t)

\int \frac{3}{x ^{2} - x} d x

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

What is (\partial)/(\partial x)(sqrt(x^2+8xy+y^2)) ?
The answer to (\partial)/(\partial x)(sqrt(x^2+8xy+y^2)) is (x+4y)/(sqrt(x^2+8xy+y^2))