What is (\partial)/(\partial x)(sin(7x^4y-7xy^4)) ?

The answer to (\partial)/(\partial x)(sin(7x^4y-7xy^4)) is cos(7x^4y-7xy^4)(28yx^3-7y^4)

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(\partial)/(\partial x)(sin(7x^4y-7xy^4))

\frac{\partial}{\partial x} (sin (7 x^{4} y - 7 x y^{4}))

cos (7 x^{4} y - 7 x y^{4}) (28 y x^{3} - 7 y^{4})

Solution steps

\frac{\partial}{\partial x} (sin (7 x^{4} y - 7 x y^{4}))

Treat

y

as a constant

Apply the chain rule:

cos (7 x^{4} y - 7 x y^{4}) \frac{\partial}{\partial x} (7 x^{4} y - 7 x y^{4})

= cos (7 x^{4} y - 7 x y^{4}) \frac{\partial}{\partial x} (7 x^{4} y - 7 x y^{4})

\frac{\partial}{\partial x} (7 x^{4} y - 7 x y^{4}) = 28 y x^{3} - 7 y^{4}

= cos (7 x^{4} y - 7 x y^{4}) (28 y x^{3} - 7 y^{4})

y^{^{''}} - 10 y^{^{'}} + 34 y = 0

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

\frac{s - 1}{s ( s ^{2} + s + 1 )}

\frac{\partial}{\partial x} (e^{e^{x y}})

x ∣ x - 2 ∣

What is (\partial)/(\partial x)(sin(7x^4y-7xy^4)) ?
The answer to (\partial)/(\partial x)(sin(7x^4y-7xy^4)) is cos(7x^4y-7xy^4)(28yx^3-7y^4)