What is the derivative of x^2+y^4 ?

The derivative of x^2+y^4 is 2x+4y^3(dy)/(dx)

What is the first derivative of x^2+y^4 ?

The first derivative of x^2+y^4 is 2x+4y^3(dy)/(dx)

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

2 x + 4 y^{3} \frac{d y}{d x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 2 x + 4 y^{3} \frac{d y}{d x}

f (x) = \frac{2 x - 1}{x + 7}, at x = 2

f (x) = \frac{1}{2 x}, at x = 100

sin (e^{- 2 q})

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

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

What is the derivative of x^2+y^4 ?
The derivative of x^2+y^4 is 2x+4y^3(dy)/(dx)
What is the first derivative of x^2+y^4 ?
The first derivative of x^2+y^4 is 2x+4y^3(dy)/(dx)