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

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

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

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

8 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (3 y^{4}) = 0

= 8 x + 0

Simplify

= 8 x

x \to 1 - lim (\frac{1}{x} - 1)

y^{^{''}} + 9 y = 3 sec (3 t)

f (x) = cos (x^{2}) \cdot 2 x

y = x + 1, y = 9 - x^{2}, x = - 1, x = 2

\int (sin^{4} (x) cos^{2} (x)) d x

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