Shirley Thompson

2021-12-30

Find all the second partial derivatives. $v=\frac{xy}{x-y}$

lovagwb

$v=\frac{xy}{x-y}$
Partially differentiate with respect to x (use the quotient rule)
${v}_{x}=\frac{\left[d\frac{xy}{dx}\right]\cdot \left(x-y\right)-\left(xy\right)\cdot \left[d\frac{x-y}{dx}\right]}{{\left(x-y\right)}^{2}}$
$=\frac{y\left(x-y\right)-\left(xy\right)\cdot \left(1-0\right)}{{\left(x-y\right)}^{2}}$
$=\frac{xy-{y}^{2}-xy}{{\left(x-y\right)}^{2}}=\frac{-{y}^{2}}{{\left(x-y\right)}^{2}}$
Partially differentiate with respect to y (use the quotient rule)
${v}_{y}=\frac{\left[d\frac{xy}{dy}\right]\cdot \left(x-y\right)-\left(xy\right)\cdot \left[d\frac{x-y}{dy}\right]}{{\left(x-y\right)}^{2}}$
$=\frac{\left(x\right)\left(x-y\right)-\left(xy\right)\cdot \left(0-1\right)}{{\left(x-y\right)}^{2}}$
$=\frac{{x}^{2}-xy+xy}{{\left(x-y\right)}^{2}}=\frac{{x}^{2}}{{\left(x-y\right)}^{2}}$
Partially differentiate ${v}_{x}$ with respect to x
${v}_{×}=\frac{\left[d\frac{-{y}^{2}}{dx}\right]\cdot {\left(x-y\right)}^{2}-\left(-{y}^{2}\right)\cdot \left[d\frac{{\left(x-y\right)}^{2}}{dx}\right]}{{\left(x-y\right)}^{4}}$
$=\frac{0\left(x-y\right)-\left(-{y}^{2}\right)\cdot 2\left(x-y\right)\cdot \left(1-0\right)}{{\left(x-y\right)}^{4}}$
Cancel $\left(x-y\right)$ from both the numerator and the denominator
$=\frac{2{y}^{2}}{{\left(x-y\right)}^{3}}$
Note that: If $w=\frac{yx}{y-x}$, then we can write ${w}_{y}y=\frac{2{x}^{2}}{{\left(y-x\right)}^{3}}$
But $w=-v$, therefore

ol3i4c5s4hr

Initial equation for 8e
$T={e}^{-2r}\mathrm{cos}\theta$
Find first partial derivative wrt. r
${T}_{r}=-2{e}^{-2r}\mathrm{cos}\theta$
Find second partial derivative wrt. r
${T}_{rr}=4{e}^{-2r}\mathrm{cos}\theta$
Find partial derivative wrt. theta
${T}_{\theta }=-{e}^{-2r}\mathrm{sin}\theta$
Find second partial derivative wrt theta
${T}_{\theta \theta }=-{e}^{-2r}\mathrm{cos}\theta$
Find the second partial derivative wrt the alternative variable
${T}_{r\theta }={T}_{\theta r}=2{e}^{-2r}\mathrm{sin}\theta$
Result:

karton

$v=\frac{xy}{x-y}$
Find each first partial derivative, using the quotient rule.
$\begin{array}{}{v}_{x}=\frac{y\left(x-y\right)-xy}{\left(x-y{\right)}^{2}}=\frac{-{y}^{2}}{\left(x-y{\right)}^{2}}\\ {v}_{y}=\frac{x\left(x-y\right)+xy}{\left(x-y{\right)}^{2}}=\frac{{x}^{2}}{\left(x-y{\right)}^{2}}\end{array}$
Find each second partial derivative, using the quotient rule.
$\begin{array}{}{v}_{xx}=\frac{2{y}^{2}}{\left(x-y{\right)}^{3}}\\ {v}_{xy}=\frac{-2y\left(x-y{\right)}^{2}-\left(-2\right)\left(x-y\right)\left(-y{\right)}^{2}}{\left(x-y{\right)}^{4}}=\frac{-2y\left(x-y\right)-2{y}^{2}}{\left(x-y{\right)}^{3}}=\frac{-2xy}{\left(x-y{\right)}^{3}}\\ {v}_{yx}=\frac{2x\left(x-y{\right)}^{2}-2\left(x-y\right){x}^{2}}{\left(x-y{\right)}^{4}}=\frac{2x\left(x-y\right)-2{x}^{2}}{\left(x-y{\right)}^{3}}=\frac{-2xy}{\left(x-y{\right)}^{3}}\\ {v}_{yy}=\frac{2{x}^{2}}{\left(x-y{\right)}^{3}}\end{array}$

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