if xa5+yb5=5, then dydxat(a,b) is

A
$\frac{a}{b}$
B
$\frac{- a}{b}$
C
$\frac{b}{a}$
D
$\frac{- b}{a}$

Solution:

$\begin{array}{l} 5. {(\frac{x}{a})}^{4} \frac{1}{a} + 5 {(\frac{y}{b})}^{4} \frac{1}{b} . y^{1} = 0 \\ 5 \frac{y^{1}}{b} . \frac{y^{4}}{b^{4}} = \frac{- 5}{a} . \frac{x^{4}}{a^{4}} \\ {\frac{d y}{d x}}_{_{(a, b)}} = y^{1} = \frac{- b^{5} a^{4}}{a^{5} b^{4}} = \frac{- b}{a} \end{array}$

if ${(\frac{x}{a})}^{5} + {(\frac{y}{b})}^{5} = 5$ , then ${\frac{d y}{d x}}_{at (a, b)}$ is

Solution:

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