What is (\partial)/(\partial x)(6x^2+y^2-7y) ?

The answer to (\partial)/(\partial x)(6x^2+y^2-7y) is 12x

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(\partial)/(\partial x)(6x^2+y^2-7y)

\frac{\partial}{\partial x} (6 x^{2} + y^{2} - 7 y)

12 x

Solution steps

\frac{\partial}{\partial x} (6 x^{2} + y^{2} - 7 y)

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} (y^{2}) - \frac{\partial}{\partial x} (7 y)

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

\frac{\partial}{\partial x} (y^{2}) = 0

\frac{\partial}{\partial x} (7 y) = 0

= 12 x + 0 - 0

Simplify

= 12 x

e^{- 6 s}

\frac{\partial}{\partial x} (56 y e^{8 x})

\int (8 + x) (2 x - 8) d x

(ln (x + 1 + x^{2}))^{^{'}}

\int \frac{( y + 1 )}{y} d y

What is (\partial)/(\partial x)(6x^2+y^2-7y) ?
The answer to (\partial)/(\partial x)(6x^2+y^2-7y) is 12x