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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

y = (4 x^{6} + 7)^{5}

9 x - 6 x

\frac{\partial}{\partial x} (2 cos (2 x))

\int \frac{5 x + 3}{x ^{2} + 4 x + 10} d x

\frac{d}{d x} (cos^{3} (\frac{x}{3}))

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