2.4 Q-1

Question Statement

Find $\frac{dy}{dx}$ by making suitable substitutions for the following cases:

1. $y = \sqrt{\frac{1-x}{1+x}}$
2. $y = \sqrt{x + \sqrt{x}}$
3. $y = x \sqrt{\frac{a+x}{a-x}}$
4. $y = (3x^2 - 2x + 7)^6$
5. $y = \sqrt{\frac{a^2 + x^2}{a^2 - x^2}}$

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Background and Explanation

To solve these problems, the chain rule is used extensively. The chain rule states:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx},$$

where intermediate substitution is made for $u$, simplifying differentiation. Knowledge of basic differentiation rules, square root derivatives, and rational function derivatives is required.

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Solution

Case 1: $y = \sqrt{\frac{1-x}{1+x}}$

1. Let $u = \frac{1-x}{1+x}$, so $y = u^{1/2}$.
2. Differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = \frac{-1(1+x) - (1-x)(1)}{(1+x)^2} = \frac{-2}{(1+x)^2}.$$

3. Differentiate $y$ with respect to $u$:

$$\frac{dy}{du} = \frac{1}{2} u^{-1/2} = \frac{1}{2\sqrt{u}}.$$

4. By the chain rule:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{\frac{1-x}{1+x}}} \cdot \frac{-2}{(1+x)^2}.$$

5. Simplify using $\sqrt{u} = \sqrt{\frac{1-x}{1+x}}$:

$$\frac{dy}{dx} = \frac{-1}{\sqrt{1-x}(1+x)^{3/2}}.$$

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Case 2: $y = \sqrt{x + \sqrt{x}}$

1. Let $u = x + \sqrt{x}$, so $y = \sqrt{u} = u^{1/2}$.
2. Differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = 1 + \frac{1}{2\sqrt{x}}.$$

3. Differentiate $y$ with respect to $u$:

$$\frac{dy}{du} = \frac{1}{2\sqrt{u}}.$$

4. By the chain rule:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{x+\sqrt{x}}} \cdot \left(1 + \frac{1}{2\sqrt{x}}\right).$$

5. Simplify:

$$\frac{dy}{dx} = \frac{1 + \frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}} = \frac{2\sqrt{x}+1}{4\sqrt{x}\sqrt{x+\sqrt{x}}}.$$

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Case 3: $y = x \sqrt{\frac{a+x}{a-x}}$

1. Let $u = \frac{a+x}{a-x}$, so $y = x\sqrt{u}$.
2. Differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = \frac{(a-x)(1) - (a+x)(-1)}{(a-x)^2} = \frac{2a}{(a-x)^2}.$$

3. Differentiate $y$ with respect to $x$ using the product rule:

$$\frac{dy}{dx} = \sqrt{u} \cdot \frac{dx}{dx} + x \cdot \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}.$$

4. Substitute values:

$$\frac{dy}{dx} = \sqrt{\frac{a+x}{a-x}} + x \cdot \frac{1}{2\sqrt{\frac{a+x}{a-x}}} \cdot \frac{2a}{(a-x)^2}.$$

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Case 4: $y = (3x^2 - 2x + 7)^6$

1. Let $u = 3x^2 - 2x + 7$, so $y = u^6$.
2. Differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = 6x - 2.$$

3. Differentiate $y$ with respect to $u$:

$$\frac{dy}{du} = 6u^5.$$

4. By the chain rule:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 6(3x^2 - 2x + 7)^5 \cdot (6x - 2).$$

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Case 5: $y = \sqrt{\frac{a^2+x^2}{a^2-x^2}}$

1. Let $u = \frac{a^2+x^2}{a^2-x^2}$, so $y = u^{1/2}$.
2. Differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = \frac{(a^2-x^2)(2x) - (a^2+x^2)(-2x)}{(a^2-x^2)^2} = \frac{4a^2x}{(a^2-x^2)^2}.$$

3. Differentiate $y$ with respect to $u$:

$$\frac{dy}{du} = \frac{1}{2} u^{-1/2}.$$

4. By the chain rule:

$$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{a^2-x^2}{a^2+x^2}} \cdot \frac{4a^2x}{(a^2-x^2)^2}.$$

5. Simplify:

$$\frac{dy}{dx} = \frac{2a^2x}{\sqrt{a^2+x^2}(a^2-x^2)^{3/2}}.$$

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Key Formulas or Methods Used

1. Chain Rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
2. Product Rule: $\frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$.
3. Square Root Derivatives: $\frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$.

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Summary of Steps

1. Identify a substitution to simplify the function.
2. Differentiate the substitution $u$ with respect to $x$.
3. Apply the chain rule and substitute back the original expressions.
4. Simplify to the final derivative.