# Question #b2c2c

Oct 5, 2017

(a) $y ' = - \frac{{y}^{2}}{{x}^{2}}$, (b) $y ' = - \frac{4}{{\left(7 x - 2\right)}^{2}}$, and (c) $y ' = - \frac{{\left(\frac{2 x}{7 x - 2}\right)}^{2}}{{x}^{2}} = - \frac{4}{{\left(7 x - 2\right)}^{2}}$

#### Explanation:

(a) Assume that $\frac{2}{x} + \frac{2}{y} = 7$, or $2 {x}^{- 1} + 2 {y}^{- 1} = 7$, implicitly defines $y$ as a function of $x$ and differentiate both sides of this equation with respect to $x$ to get $- 2 {x}^{- 2} - 2 {y}^{- 2} \cdot y ' = 0$ (the Chain Rule is used here in the second term).

This can be rearranged to say that $- \frac{y '}{{y}^{2}} = \frac{1}{{x}^{2}}$, or $y ' = - \frac{{y}^{2}}{{x}^{2}}$.

(b) Do algebra to solve $\frac{2}{x} + \frac{2}{y} = 7$ as follows: $\frac{2}{y} = 7 - \frac{2}{x} = \frac{7 x - 2}{x} = \implies \frac{y}{2} = \frac{x}{7 x - 2} = \implies y = \frac{2 x}{7 x - 2}$.

Now differentiate with the Quotient Rule:

$y ' = \frac{\left(7 x - 2\right) \cdot 2 - \left(2 x\right) \cdot 7}{{\left(7 x - 2\right)}^{2}}$

$= \frac{14 x - 4 - 14 x}{{\left(7 x - 2\right)}^{2}} = - \frac{4}{{\left(7 x - 2\right)}^{2}}$

(c) Finally, substitute $y = \frac{2 x}{7 x - 2}$ into the equation $y ' = - \frac{{y}^{2}}{{x}^{2}}$ and simplify:

$y ' = - \frac{{\left(\frac{2 x}{7 x - 2}\right)}^{2}}{{x}^{2}} = - \frac{4 {x}^{2}}{{\left(7 x - 2\right)}^{2}} \cdot \frac{1}{{x}^{2}} = - \frac{4}{{\left(7 x - 2\right)}^{2}}$, which is the same as the answer from part (b).