What is (\partial)/(\partial x)(x^2-kxe^{3y}-3y^2) ?

The answer to (\partial)/(\partial x)(x^2-kxe^{3y}-3y^2) is 2x-ke^{3y}

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(\partial)/(\partial x)(x^2-kxe^{3y}-3y^2)

\frac{\partial}{\partial x} (x^{2} - k x e^{3 y} - 3 y^{2})

2 x - k e^{3 y}

Solution steps

\frac{\partial}{\partial x} (x^{2} - k x e^{3 y} - 3 y^{2})

Treat

k, y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (k x e^{3 y}) = k e^{3 y}

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

= 2 x - k e^{3 y} - 0

Simplify

= 2 x - k e^{3 y}

f (x) = x^{2} - 5, (2, - 1)

\frac{( 5 x - 4 )}{x - 1}

\frac{d}{d x} (- s t)

\int \frac{1}{x ^{2} x} d x

3 ln (x)

What is (\partial)/(\partial x)(x^2-kxe^{3y}-3y^2) ?
The answer to (\partial)/(\partial x)(x^2-kxe^{3y}-3y^2) is 2x-ke^{3y}