Question

If $\sqrt{\frac{v}{\mu }}+\sqrt{\frac{\mu }{v}}=6,$ then $\frac{dv}{d\mu }=$

• A. $\frac{17\mu -v}{\mu -17v}$
• B. $\frac{\mu -17v}{17\mu -v}$
• C. $\frac{17\mu +v}{\mu -17v}$
• D. $\frac{\mu +17v}{17\mu -v}$

Answer 1

The correct option is B $\frac{\mu -17v}{17\mu -v}$

Given $\sqrt{\frac{u}{\mu }}+\sqrt{\frac{\mu }{u}}=6$

Squaring on both sides

$\Rightarrow {[\sqrt{\frac{u}{\mu }}+\sqrt{\frac{\mu }{\mu }}]}^2=6^2$

$\Rightarrow {\left(\sqrt{\frac{u}{\mu }}\right)}^2+{\left(\sqrt{\frac{\mu }{u}}\right)}^2+2\sqrt{\frac{u}{\mu }}\times \sqrt{\frac{\mu }{u}}=36$

$\Rightarrow \frac{u}{\mu }+\frac{\mu }{u}+2=36$

$\Rightarrow \frac{u}{\mu }+\frac{\mu }{u}=34$

$\Rightarrow \frac{u^2+{\mu }^2}{u\mu }=34$

$\Rightarrow u^2+{\mu }^2=34u\mu$

Differentiating both sides with respect to $\left(\mu \right)$

$\Rightarrow 2u\frac{du}{d\mu }+2\mu =34(u+\mu \frac{du}{d\mu })$

$\Rightarrow u.\frac{d\mu }{d\mu }+\mu =17u+17\mu \frac{du}{d\mu }$

$\Rightarrow (17\mu -u)\frac{du}{d\mu }=17u-\mu$

$\Rightarrow \frac{du}{d\mu }=\frac{17u-\mu }{17\mu -u}$