What is (\partial)/(\partial y)(xy^6+x^2+y^4) ?

The answer to (\partial)/(\partial y)(xy^6+x^2+y^4) is 6xy^5+4y^3

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(\partial)/(\partial y)(xy^6+x^2+y^4)

\frac{\partial}{\partial y} (x y^{6} + x^{2} + y^{4})

6 x y^{5} + 4 y^{3}

Solution steps

\frac{\partial}{\partial y} (x y^{6} + x^{2} + y^{4})

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (x y^{6}) + \frac{\partial}{\partial y} (x^{2}) + \frac{\partial}{\partial y} (y^{4})

\frac{\partial}{\partial y} (x y^{6}) = 6 x y^{5}

\frac{\partial}{\partial y} (x^{2}) = 0

\frac{\partial}{\partial y} (y^{4}) = 4 y^{3}

= 6 x y^{5} + 0 + 4 y^{3}

Simplify

= 6 x y^{5} + 4 y^{3}

\frac{\partial}{\partial z} (arctan (\frac{x}{z}))

- 2 x^{2} + 50 x + 25.247

\frac{d}{d x} (x^{3} + 10)

\int \frac{ln ( x + 1 )}{( x + 1 ) ^{3}} d x

y^{^{'}} = (x + 1) \cdot (y + 1)

What is (\partial)/(\partial y)(xy^6+x^2+y^4) ?
The answer to (\partial)/(\partial y)(xy^6+x^2+y^4) is 6xy^5+4y^3