What is (\partial)/(\partial z)(xyarcsin(yz)) ?

The answer to (\partial)/(\partial z)(xyarcsin(yz)) is (xy^2)/(sqrt(1-y^2z^2))

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(\partial)/(\partial z)(xyarcsin(yz))

\frac{\partial}{\partial z} (x y arcsin (yz))

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

Solution steps

\frac{\partial}{\partial z} (x y arcsin (yz))

Treat

x, y

as constants

Take the constant out:

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

= x y \frac{\partial}{\partial z} (arcsin (yz))

Apply the chain rule:

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

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

\frac{\partial}{\partial z} (yz) = y

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

Simplify

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

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

\frac{d}{d x} (3 + \frac{1}{2} cosh (2 x))

f (x) = x^{2} - 9 x + 17, at x = 4

x^{3} - 3 x + 3

\frac{8 x}{x ^{2} + 3}

n = 0 \sum \infty (\frac{1}{7})^{n}

What is (\partial)/(\partial z)(xyarcsin(yz)) ?
The answer to (\partial)/(\partial z)(xyarcsin(yz)) is (xy^2)/(sqrt(1-y^2z^2))