What is the derivative of 3x^2+2y^4 ?

The derivative of 3x^2+2y^4 is 6x+8y^3(dy)/(dx)

What is the first derivative of 3x^2+2y^4 ?

The first derivative of 3x^2+2y^4 is 6x+8y^3(dy)/(dx)

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

6 x + 8 y^{3} \frac{d y}{d x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (3 x^{2}) = 6 x

\frac{d}{d x} (2 y^{4}) = 8 y^{3} \frac{d y}{d x}

= 6 x + 8 y^{3} \frac{d y}{d x}

\frac{d}{d y} (e^{- x y})

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

\frac{d y}{d x} = (\frac{1}{sec ^{2} ( y )})

sin (t + \frac{π}{3})

f (x) = - 3

What is the derivative of 3x^2+2y^4 ?
The derivative of 3x^2+2y^4 is 6x+8y^3(dy)/(dx)
What is the first derivative of 3x^2+2y^4 ?
The first derivative of 3x^2+2y^4 is 6x+8y^3(dy)/(dx)