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Differential Calculus

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Question : 68 of 69

Marks: +1, -0

If

y=\sqrt{\frac{1-x}{1+x}}

(1-x^{2})\frac{d y}{d x}=

y
$-x$
$-y$
0

Solution:

Concept:

Differentiate a composite function using logarithmic differentiation.

Explanation:

Let

y = \sqrt{\frac{1-x}{1+x}} = \left( \frac{1-x}{1+x} \right)^{1/2}

.

Take natural log:

\ln y = \frac{1}{2} [\ln(1-x) - \ln(1+x)]

.

Differentiate both sides:

\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left( -\frac{1}{1-x} - \frac{1}{1+x} \right)

.

Combine the fractions inside:

-\frac{1}{1-x} - \frac{1}{1+x} = -\left( \frac{1}{1-x} + \frac{1}{1+x} \right) = -\frac{2}{1-x^2}

.

Thus

\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left( -\frac{2}{1-x^2} \right) = -\frac{1}{1-x^2}

.

Multiply both sides by

y

:

\frac{dy}{dx} = -\frac{y}{1-x^2}

.

Multiply both sides by

(1-x^2)

:

(1-x^2) \frac{dy}{dx} = -y

.

Answer:

(1-x^{2})\frac{dy}{dx} = -y

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