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

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

Popular >

(\partial)/(\partial x)(sqrt(x^2-y^2-z^2))

\frac{\partial}{\partial x} (x^{2} - y^{2} - z^{2})

\frac{x}{x ^{2} - y ^{2} - z ^{2}}

Solution steps

\frac{\partial}{\partial x} (x^{2} - y^{2} - z^{2})

Treat

y, z

as constants

Apply the chain rule:

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

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

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

= \frac{1}{2 x ^{2} - y ^{2} - z ^{2}} \cdot 2 x

Simplify

\frac{1}{2 x ^{2} - y ^{2} - z ^{2}} \cdot 2 x : \frac{x}{x ^{2} - y ^{2} - z ^{2}}

= \frac{x}{x ^{2} - y ^{2} - z ^{2}}

f (x) = 2 x^{2} - 1

x \to 0 + lim ((1 + x^{2})^{\frac{1}{x}})

y^{^{''}} + 16 y = cos (4 x) + sin (4 x)

\int + \frac{x ^{2}}{1 - 2 x ^{3}} d x

n = 0 \sum \infty \frac{( n - 1 )}{n}

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