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Differential Calculus

Question : 33 of 69

Marks: +1, -0

y^{3} \cdot x^{5} = (x+y)^{8},

\; \frac{dy}{dx}

Solution:

Concept:

Use logarithmic differentiation to simplify the equation before differentiating.

Explanation:

Take natural logarithms on both sides:

\ln(y^3) + \ln(x^5) = \ln((x+y)^8)

This gives

3\ln y + 5\ln x = 8\ln(x+y)

Differentiate with respect to

x

\frac{3}{y}\frac{dy}{dx} + \frac{5}{x} = \frac{8}{x+y}\left(1 + \frac{dy}{dx}\right)

Rearrange terms:

\frac{3}{y}y' - \frac{8}{x+y}y' = \frac{8}{x+y} - \frac{5}{x}

Factor

y'

y'\left(\frac{3}{y} - \frac{8}{x+y}\right) = \frac{8}{x+y} - \frac{5}{x}

Simplify each side: numerator =

\frac{3x-5y}{x(x+y)}

, denominator =

\frac{3x-5y}{y(x+y)}

Thus

y' = \frac{(3x-5y)/(x(x+y))}{(3x-5y)/(y(x+y))} = \frac{y}{x}

Answer:

\frac{dy}{dx} = \frac{y}{x}

, which corresponds to option A.

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