3 years ago

(m^-4/3)^1/4 divided by m^0 n^3/4(m^-3/2 n^2)

1 answer:

6 0

$\frac{(m^{-\frac{4}{3}})^{\frac{1}{4}}}{m^{0}(m^{-\frac{3}{2}}n^{2})} = \frac{m^{-\frac{1}{3}}}{m^{-\frac{3}{2}}n^{2}} = \frac{m^{\frac{3}{2}}}{m^{\frac{1}{3}}n^{2}} = \frac{m^{\frac{7}{6}}}{n^{2}} = \frac{\sqrt[6]{m^{7}}}{n^{2}} = \frac{\sqrt[6]{m^{6} * m}}{n^{2}} = \frac{\sqrt[6]{m^{6}}\sqrt[6]{m}}{n^{2}} = \frac{m\sqrt[6]{m}}{n^{2}}$

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Find an equation of the tangent line to the curve 2(x2+y2)2=25(x2−y2) (a lemniscate) at the point (−3,1). An equation of the tan

valina [46]

$2(x^2+y^2)^2=25(x^2-y^2)$

Let $y=y(x)$ , so that differentiating both sides wrt $x$ gives

$4(x^2+y^2)\left(2x+2y\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(2x-2y\dfrac{\mathrm dy}{\mathrm dx}\right)$

If $x=-3$ and $y=1$ , the above reduces to

$40\left(-6+2\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(-6-2\dfrac{\mathrm dy}{\mathrm dx}\right)\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac9{13}$