What is (\partial)/(\partial x)(6x^2+5x^3y^4-10y^5) ?

The answer to (\partial)/(\partial x)(6x^2+5x^3y^4-10y^5) is 15x^2y^4+12x

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(\partial)/(\partial x)(6x^2+5x^3y^4-10y^5)

\frac{\partial}{\partial x} (6 x^{2} + 5 x^{3} y^{4} - 10 y^{5})

15 x^{2} y^{4} + 12 x

Solution steps

\frac{\partial}{\partial x} (6 x^{2} + 5 x^{3} y^{4} - 10 y^{5})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (6 x^{2}) + \frac{\partial}{\partial x} (5 x^{3} y^{4}) - \frac{\partial}{\partial x} (10 y^{5})

\frac{\partial}{\partial x} (6 x^{2}) = 12 x

\frac{\partial}{\partial x} (5 x^{3} y^{4}) = 15 y^{4} x^{2}

\frac{\partial}{\partial x} (10 y^{5}) = 0

= 12 x + 15 y^{4} x^{2} - 0

Simplify

= 15 x^{2} y^{4} + 12 x

\int \frac{3}{x ^{2} x ^{2} - 9} d x

\int \frac{x + 2}{2 x - 1} d x

y = e^{3 x}, y = e^{4 x}, x = 1

x = 7 - 7 y^{2}, x = 7 y^{2} - 7

2 t y^{2} d t + 2 t^{2} y d y = 0

What is (\partial)/(\partial x)(6x^2+5x^3y^4-10y^5) ?
The answer to (\partial)/(\partial x)(6x^2+5x^3y^4-10y^5) is 15x^2y^4+12x