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

The answer to (\partial)/(\partial y)(tan(x^2-y^2)) is -2ysec^2(x^2-y^2)

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(\partial)/(\partial y)(tan(x^2-y^2))

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

- 2 y sec^{2} (x^{2} - y^{2})

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

= sec^{2} (x^{2} - y^{2}) (- 2 y)

Simplify

= - 2 y sec^{2} (x^{2} - y^{2})

x y^{^{'}} + y = 29 x

\int 4 - 4 x^{2} d x

2 x^{4}

\frac{d}{d x} (\frac{x ^{3}}{8})

x \to - 4 + lim (\frac{2}{( x + 4 ) ^{3}})

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