What is (\partial)/(\partial x)(x^5+y^5-5x^3y^3) ?

The answer to (\partial)/(\partial x)(x^5+y^5-5x^3y^3) is 5x^4-15x^2y^3

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(\partial)/(\partial x)(x^5+y^5-5x^3y^3)

\frac{\partial}{\partial x} (x^{5} + y^{5} - 5 x^{3} y^{3})

5 x^{4} - 15 x^{2} y^{3}

Solution steps

\frac{\partial}{\partial x} (x^{5} + y^{5} - 5 x^{3} y^{3})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{5}) + \frac{\partial}{\partial x} (y^{5}) - \frac{\partial}{\partial x} (5 x^{3} y^{3})

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

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

\frac{\partial}{\partial x} (5 x^{3} y^{3}) = 15 y^{3} x^{2}

= 5 x^{4} + 0 - 15 y^{3} x^{2}

Simplify

= 5 x^{4} - 15 x^{2} y^{3}

x \to 1 lim (\frac{x ^{2} - 1}{x + 2})

f (x) = 9 x^{2}, at x = - 2

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

\int 3 x^{- 4} (2 + 4 x^{- 3})^{- 7} d x

f^{^{'}} (t) - f (t) = - 4 t

What is (\partial)/(\partial x)(x^5+y^5-5x^3y^3) ?
The answer to (\partial)/(\partial x)(x^5+y^5-5x^3y^3) is 5x^4-15x^2y^3