What is (\partial)/(\partial x)(25-(x^2+3y^2)) ?

The answer to (\partial)/(\partial x)(25-(x^2+3y^2)) is -2x

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(\partial)/(\partial x)(25-(x^2+3y^2))

\frac{\partial}{\partial x} (25 - (x^{2} + 3 y^{2}))

- 2 x

Solution steps

\frac{\partial}{\partial x} (25 - (x^{2} + 3 y^{2}))

Simplify

25 - (x^{2} + 3 y^{2}) : 25 - x^{2} - 3 y^{2}

= \frac{\partial}{\partial x} (25 - x^{2} - 3 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (25) - \frac{\partial}{\partial x} (x^{2}) - \frac{\partial}{\partial x} (3 y^{2})

\frac{\partial}{\partial x} (25) = 0

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

\frac{\partial}{\partial x} (3 y^{2}) = 0

= 0 - 2 x - 0

Simplify

= - 2 x

\int 2 y + z d x

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 5 y = 2 sin (x)

\frac{\partial}{\partial x} (2 x yz)

y = \frac{3 x}{x}

\int \frac{ln ^{6} ( x )}{x} d x

What is (\partial)/(\partial x)(25-(x^2+3y^2)) ?
The answer to (\partial)/(\partial x)(25-(x^2+3y^2)) is -2x