What is the derivative of 3x^2y^2+1 ?

The derivative of 3x^2y^2+1 is 3(2x^2y(dy)/(dx)+2xy^2)

What is the first derivative of 3x^2y^2+1 ?

The first derivative of 3x^2y^2+1 is 3(2x^2y(dy)/(dx)+2xy^2)

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (1) = 0

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

Simplify

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

\int e^{0.05 x} d x

n = 1 \sum \infty 1 1^{n} x^{n} n!

\frac{1}{1 + x ^{3}}

\frac{( 11 ln ( x ) )}{( 2 x + 4 )}

\frac{d y}{d x} = \frac{- x}{y}

What is the derivative of 3x^2y^2+1 ?
The derivative of 3x^2y^2+1 is 3(2x^2y(dy)/(dx)+2xy^2)
What is the first derivative of 3x^2y^2+1 ?
The first derivative of 3x^2y^2+1 is 3(2x^2y(dy)/(dx)+2xy^2)