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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

\frac{\partial}{\partial x} (20 x + 80 y - x^{2} - 2 y^{2})

\frac{d}{d x} (sin^{7} (2 x))

\frac{d}{d x} (- 10 e^{- 2 x})

y = \frac{6 x}{7 - tan ( x )}

\int \frac{1}{1 + 2 y} d y

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