Question

plzz solve fast..the circled sum. ..derivative 2nd order

Answer 1

Hope u like my process
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${x}^{p} {y}^{q} = (x + y) ^{p + q}$
Taking log on both sides we get
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p.log(x) + q.log(y) = (p+q).log(x +y)

Differentiating both sides by x, we get
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$\frac{p}{x} + \frac{q}{y} \times \frac{dy}{dx} = (p + q) \times \frac{1}{x + y} (1 + \frac{dy}{dx} ) \\ \\ or. \: \: \frac{q}{y} \times \frac{dy}{dx} = \frac{p + q}{x + y} + \frac{p + q}{x + y} \times \frac{dy}{dx} \\ \\ or. \: \: \frac{dy}{dx} - \frac{y}{q} \times \frac{p + q}{x + y} \times \frac{dy}{dx} = \frac{y(p + q)}{q(x + y)} \\ \\ or. \: \: \frac{dy}{dx} \times (1 - \frac{y(p + q)}{q(x + y)} ) = \frac{y(p + q)}{q(x + y)} \\ \\ or. \: \: \frac{dy}{dx} \times \frac{(q(x + y) - y(p + q))}{q(x + y)} \times q(x + y) = y(p + q) \\ \\ or. \: \: \frac{dy}{dx} = \frac{y(p + q)}{(q(x + y) - y(p + q))} \\ \\ or. \: \: \frac{dy}{dx} = \frac{1}{ \frac{q(x + y)}{y(p + q)} - 1}$
Again Differentiating in both sides we get!!
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Hope this is ur required answer

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