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

The derivative of 3sqrt(x^2+y^2) is (3(x+y\frac{dy)/(dx))}{sqrt(x^2+y^2)}

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

The first derivative of 3sqrt(x^2+y^2) is (3(x+y\frac{dy)/(dx))}{sqrt(x^2+y^2)}

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

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

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

Simplify

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

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

\frac{\partial}{\partial x} (- 7 x^{2} e^{6 x})

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

x = 3 y^{2}, x = 0, [0, 2]

\int \frac{5 x ^{4} + 6}{( x ^{5} + 6 x ) ^{2}} d x

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

What is the derivative of 3sqrt(x^2+y^2) ?
The derivative of 3sqrt(x^2+y^2) is (3(x+y\frac{dy)/(dx))}{sqrt(x^2+y^2)}
What is the first derivative of 3sqrt(x^2+y^2) ?
The first derivative of 3sqrt(x^2+y^2) is (3(x+y\frac{dy)/(dx))}{sqrt(x^2+y^2)}