What is (\partial)/(\partial x)(8x^5y^3-9x^4y^6) ?

The answer to (\partial)/(\partial x)(8x^5y^3-9x^4y^6) is 40y^3x^4-36y^6x^3

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(\partial)/(\partial x)(8x^5y^3-9x^4y^6)

\frac{\partial}{\partial x} (8 x^{5} y^{3} - 9 x^{4} y^{6})

40 y^{3} x^{4} - 36 y^{6} x^{3}

Solution steps

\frac{\partial}{\partial x} (8 x^{5} y^{3} - 9 x^{4} y^{6})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (8 x^{5} y^{3}) - \frac{\partial}{\partial x} (9 x^{4} y^{6})

\frac{\partial}{\partial x} (8 x^{5} y^{3}) = 40 y^{3} x^{4}

\frac{\partial}{\partial x} (9 x^{4} y^{6}) = 36 y^{6} x^{3}

= 40 y^{3} x^{4} - 36 y^{6} x^{3}

\frac{d}{d x} (\frac{4 x ^{3} + 1}{x ^{2} - x + 3})

\frac{d y}{d x} - \frac{y}{x} = \frac{x ^{2} + y ^{2}}{x}

f (x) = e^{\frac{1}{2} x}

\frac{d}{d x} (x + 7)

\frac{d y}{d x}, x^{2} + 5 x + 9 x y - y^{2} = 4

What is (\partial)/(\partial x)(8x^5y^3-9x^4y^6) ?
The answer to (\partial)/(\partial x)(8x^5y^3-9x^4y^6) is 40y^3x^4-36y^6x^3