Solved: "Let f(x,y)=1x2+y2. Compute ∇f(x,y)"

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Falak Kinney

Answered question

2021-09-24

Let $f (x, y) = \frac{1}{\sqrt{x^{2} + y^{2}}}$ . Compute $\nabla f (x, y)$

Answer & Explanation

aprovard

Skilled2021-09-25Added 94 answers

Let $f : R^{n} \to R$ be a differentiable function. Remember that the gradient of function f is defined in the following way:
$\nabla f = (\frac{d f}{{d x}_{1}}, \frac{d f}{{d x}_{2}}, \dots, \frac{d f}{{d x}_{n}})$
To calculate $\frac{d f}{d x}$ imagine that y=C for some constant $C \in R$ and take the derivative with respect to x. To obtain $\frac{d f}{d y}$
$\frac{d f}{d x} (x, y) = \frac{d}{d x} \frac{1}{\sqrt{x^{2} + C^{2}}} = \frac{- 1}{2} \frac{2 x}{{(x^{2} + C^{2})}^{\frac{3}{2}}}$
$= - \frac{x}{{(x^{2} + C^{2})}^{\frac{3}{2}}} = - \frac{x}{{(x^{2} + y^{2})}^{\frac{3}{2}}}$
Employing the analogous tehnique or observing the symmetry yields:
$\frac{d f}{d y} (x, y) = - \frac{y}{{(x^{2} + y^{2})}^{\frac{3}{2}}}$ Which gives
$\nabla s f (x, y) = (- \frac{x}{{(x^{2} + y^{2})}^{\frac{3}{2}}}, - \frac{y}{{(x^{2} + y^{2})}^{\frac{3}{2}}}$

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