What is (\partial)/(\partial x)(x^6-4x^2y^4+3y^2) ?

The answer to (\partial)/(\partial x)(x^6-4x^2y^4+3y^2) is 6x^5-8xy^4

(\partial)/(\partial x)(x^6-4x^2y^4+3y^2)

\frac{\partial}{\partial x} (x^{6} - 4 x^{2} y^{4} + 3 y^{2})

6 x^{5} - 8 x y^{4}

Solution steps

\frac{\partial}{\partial x} (x^{6} - 4 x^{2} y^{4} + 3 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{6}) - \frac{\partial}{\partial x} (4 x^{2} y^{4}) + \frac{\partial}{\partial x} (3 y^{2})

\frac{\partial}{\partial x} (x^{6}) = 6 x^{5}

\frac{\partial}{\partial x} (4 x^{2} y^{4}) = 8 y^{4} x

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

= 6 x^{5} - 8 y^{4} x + 0

Simplify

= 6 x^{5} - 8 x y^{4}

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

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

s im pl i f y ln (1 0^{x})

\frac{d}{d y} (x^{2} y^{4})

\frac{d y}{d t} t + 5 y = 8 t

What is (\partial)/(\partial x)(x^6-4x^2y^4+3y^2) ?
The answer to (\partial)/(\partial x)(x^6-4x^2y^4+3y^2) is 6x^5-8xy^4