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

The answer to (\partial)/(\partial x)((x+y)^5+(x-y)^5) is 10x^4+60y^2x^2+10y^4

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

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

10 x^{4} + 60 y^{2} x^{2} + 10 y^{4}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (2 x^{5}) = 10 x^{4}

\frac{\partial}{\partial x} (20 x^{3} y^{2}) = 60 y^{2} x^{2}

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

= 10 x^{4} + 60 y^{2} x^{2} + 10 y^{4}

f (x) = tan^{2} (x), at x = - \frac{π}{4}

f (x) = 13 x e^{x}

y = 3 1 + 8 x

\int \frac{1}{2 π} e^{- \frac{x ^{2}}{2}} d x

\int \frac{5 x + 12}{x ( x ^{2} + 4 )} d x

What is (\partial)/(\partial x)((x+y)^5+(x-y)^5) ?
The answer to (\partial)/(\partial x)((x+y)^5+(x-y)^5) is 10x^4+60y^2x^2+10y^4