What is (\partial)/(\partial y)(xy^8+x^2+y^4) ?

The answer to (\partial)/(\partial y)(xy^8+x^2+y^4) is 8xy^7+4y^3

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(\partial)/(\partial y)(xy^8+x^2+y^4)

\frac{\partial}{\partial y} (x y^{8} + x^{2} + y^{4})

8 x y^{7} + 4 y^{3}

Solution steps

\frac{\partial}{\partial y} (x y^{8} + 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^{8}) + \frac{\partial}{\partial y} (x^{2}) + \frac{\partial}{\partial y} (y^{4})

\frac{\partial}{\partial y} (x y^{8}) = 8 x y^{7}

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

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

= 8 x y^{7} + 0 + 4 y^{3}

Simplify

= 8 x y^{7} + 4 y^{3}

\int \frac{sin ( 9 ln ( t ) )}{t} d t

n = 1 \sum \infty arctan (6 n)

\int x^{2} x d x

\frac{d}{d t} (e^{(7 t s i n (2 t))})

e^{, at}

What is (\partial)/(\partial y)(xy^8+x^2+y^4) ?
The answer to (\partial)/(\partial y)(xy^8+x^2+y^4) is 8xy^7+4y^3