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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= 2 \cdot \frac{1}{2 x ^{2} + y ^{2}} \cdot 2 x

Simplify

2 \cdot \frac{1}{2 x ^{2} + y ^{2}} \cdot 2 x : \frac{2 x}{x ^{2} + y ^{2}}

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

\frac{d}{d x} (x^{25})

∣ x ∣ + 1

x \to 0 lim (4 \cdot x^{2})

\int 1 + cos (2 x) d x

n = 1 \sum \infty (\frac{1}{2})^{99 - 1}

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