What is (\partial)/(\partial x)(1/(sin(x^2-y^2))) ?

The answer to (\partial)/(\partial x)(1/(sin(x^2-y^2))) is -cot(x^2-y^2)csc(x^2-y^2)*2x

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(\partial)/(\partial x)(1/(sin(x^2-y^2)))

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

- cot (x^{2} - y^{2}) csc (x^{2} - y^{2}) \cdot 2 x

Solution steps

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

Simplify

\frac{1}{sin ( x ^{2} - y ^{2} )} : csc (x^{2} - y^{2})

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

= - cot (x^{2} - y^{2}) csc (x^{2} - y^{2}) \cdot 2 x

6 x

y^{^{''}} + a = 0

x y^{^{'}} + 2 y = \frac{( 3 e ^{- 3 x} )}{x}

f^{^{'}} (x) = x^{4}

\int sec^{4} (θ) d θ

What is (\partial)/(\partial x)(1/(sin(x^2-y^2))) ?
The answer to (\partial)/(\partial x)(1/(sin(x^2-y^2))) is -cot(x^2-y^2)csc(x^2-y^2)*2x