What is (\partial)/(\partial y)((x+y)^5+(x-y)^5) ?

The answer to (\partial)/(\partial y)((x+y)^5+(x-y)^5) is 40x^3y+40xy^3

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(\partial)/(\partial y)((x+y)^5+(x-y)^5)

\frac{\partial}{\partial y} ((x + y)^{5} + (x - y)^{5})

40 x^{3} y + 40 x y^{3}

Solution steps

\frac{\partial}{\partial y} ((x + y)^{5} + (x - y)^{5})

Simplify

(x + y)^{5} + (x - y)^{5} : 2 x^{5} + 20 x^{3} y^{2} + 10 x y^{4}

= \frac{\partial}{\partial y} (2 x^{5} + 20 x^{3} y^{2} + 10 x y^{4})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (2 x^{5}) + \frac{\partial}{\partial y} (20 x^{3} y^{2}) + \frac{\partial}{\partial y} (10 x y^{4})

\frac{\partial}{\partial y} (2 x^{5}) = 0

\frac{\partial}{\partial y} (20 x^{3} y^{2}) = 40 x^{3} y

\frac{\partial}{\partial y} (10 x y^{4}) = 40 x y^{3}

= 0 + 40 x^{3} y + 40 x y^{3}

Simplify

= 40 x^{3} y + 40 x y^{3}

f (x) = x^{2} + 5, (- 5, 30)

x y^{^{''}} - 2 y^{^{'}} + 4 = 0

(- 3.7) (- 4.12)

f (x) = x^{3} (3 - x)^{4}, at x = 2

x \to 100 lim (20 + 0.188 x)

What is (\partial)/(\partial y)((x+y)^5+(x-y)^5) ?
The answer to (\partial)/(\partial y)((x+y)^5+(x-y)^5) is 40x^3y+40xy^3