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

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

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

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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