What is (\partial)/(\partial x)(8y^4x^7-7y^3x^6) ?

The answer to (\partial)/(\partial x)(8y^4x^7-7y^3x^6) is 56y^4x^6-42y^3x^5

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(\partial)/(\partial x)(8y^4x^7-7y^3x^6)

\frac{\partial}{\partial x} (8 y^{4} x^{7} - 7 y^{3} x^{6})

56 y^{4} x^{6} - 42 y^{3} x^{5}

Solution steps

\frac{\partial}{\partial x} (8 y^{4} x^{7} - 7 y^{3} x^{6})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (8 y^{4} x^{7}) - \frac{\partial}{\partial x} (7 y^{3} x^{6})

\frac{\partial}{\partial x} (8 y^{4} x^{7}) = 56 y^{4} x^{6}

\frac{\partial}{\partial x} (7 y^{3} x^{6}) = 42 y^{3} x^{5}

= 56 y^{4} x^{6} - 42 y^{3} x^{5}

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

\frac{d}{d t} (e^{t} sin (2 t))

\frac{d}{d x} (\frac{2}{( x - 2 ) ^{2}})

\int \frac{4 x ^{3} - x ^{2} + 2}{x ^{4} - x ^{3} + x ^{2}} d x

4 x^{2} \frac{d y}{d x} = \frac{d y}{d x} + 4 x e^{- y}

What is (\partial)/(\partial x)(8y^4x^7-7y^3x^6) ?
The answer to (\partial)/(\partial x)(8y^4x^7-7y^3x^6) is 56y^4x^6-42y^3x^5