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

The derivative of sqrt(1+x^2+y^2) is (x+y\frac{dy)/(dx)}{sqrt(1+x^2+y^2)}

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

The first derivative of sqrt(1+x^2+y^2) is (x+y\frac{dy)/(dx)}{sqrt(1+x^2+y^2)}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{x ^{2} + 5 x + 6}{x ^{2} - 4}

\frac{2}{( 6 x ^{2} + 5 x + 1 ) ^{2}}

\int \frac{1}{aa uu} d u

\frac{d}{d x} (\frac{1}{sin ( x ^{2} )})

What is the derivative of sqrt(1+x^2+y^2) ?
The derivative of sqrt(1+x^2+y^2) is (x+y\frac{dy)/(dx)}{sqrt(1+x^2+y^2)}
What is the first derivative of sqrt(1+x^2+y^2) ?
The first derivative of sqrt(1+x^2+y^2) is (x+y\frac{dy)/(dx)}{sqrt(1+x^2+y^2)}