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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

11 x^{2} - 4 x

y^{^{'}} = \frac{( sec ( y ) )}{( x - 4 ) ^{2}}

\frac{d y}{d x} = \frac{2 - e ^{x}}{3 + 2 y}, y (0) = 0

\int_{0}^{1} (x + 1)^{2} d x

\int_{0}^{π} tan^{2} (x) d x

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