Falak Kinney

2021-09-24

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

### Answer & Explanation

aprovard

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

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