2.5 Q-4

Question Statement

Find the derivative with respect to $x$ for the following expressions:

$y = \sqrt{\frac{1+x}{1+2x}}$
$y = \sin \left( \sqrt{\frac{1+2x}{1+x}} \right)$

Background and Explanation

To solve these problems, we will apply the chain rule and quotient rule for differentiation. The chain rule is used when differentiating composite functions, and the quotient rule is applied when differentiating a ratio of two functions.

Solution

i. Differentiating $y = \sqrt{\frac{1+x}{1+2x}}$

First, we express the square root as a fractional power:

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

Now, we differentiate using the chain rule and the quotient rule. Let $u = \frac{1+x}{1+2x}$ , so that:

y = u^{1/2}

Differentiating with respect to $x$ :

\frac{d y}{d x} = \frac{1}{2} u^{-1/2} \cdot \frac{d u}{d x}

Next, we find $\frac{d u}{d x}$ using the quotient rule. For $u = \frac{1+x}{1+2x}$ , apply the quotient rule:

\frac{d u}{d x} = \frac{(1+2x) \cdot \frac{d}{d x}(1+x) - (1+x) \cdot \frac{d}{d x}(1+2x)}{(1+2x)^2}

Simplifying:

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

Substitute $\frac{d u}{d x}$ back into the expression for $\frac{d y}{d x}$ :

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

Simplifying further:

\frac{d y}{d x} = \frac{x}{2 \sqrt{(1+x)(1+2x)^3}} \quad \text{Ans}

ii. Differentiating $y = \sin \left( \sqrt{\frac{1+2x}{1+x}} \right)$

We begin by applying the chain rule. Let:

u = \sqrt{\frac{1+2x}{1+x}}

Thus, $y = \sin(u)$ , and we differentiate:

\frac{d y}{d x} = \cos(u) \cdot \frac{d u}{d x}

Now, differentiate $u = \sqrt{\frac{1+2x}{1+x}}$ . This is another composite function, so we use the chain rule again:

\frac{d u}{d x} = \frac{1}{2} \left( \frac{1+2x}{1+x} \right)^{-1/2} \cdot \frac{d}{d x} \left( \frac{1+2x}{1+x} \right)

Use the quotient rule to differentiate $\frac{1+2x}{1+x}$ :

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

Now substitute this into the expression for $\frac{d u}{d x}$ :

\frac{d u}{d x} = \frac{1}{2 \sqrt{\frac{1+2x}{1+x}}} \cdot \frac{1}{(1+x)^2}

Finally, substitute this into the derivative of $y$ :

\frac{d y}{d x} = \cos \left( \sqrt{\frac{1+2x}{1+x}} \right) \cdot \frac{1}{2 \sqrt{\frac{1+2x}{1+x}} (1+x)^3}

Simplifying:

\frac{d y}{d x} = \frac{\cos \left( \sqrt{\frac{1+2x}{1+x}} \right)}{2 \sqrt{\frac{1+2x}{1+x}} (1+x)^{3/2}} \quad \text{Ans}

Key Formulas or Methods Used

Chain Rule: Used when differentiating composite functions. If $y = f(g(x))$ , then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$ .
Quotient Rule: Used when differentiating ratios of functions. If $u(x) = \frac{f(x)}{g(x)}$ , then $\frac{du}{dx} = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2}$ .

Summary of Steps

For part i:
- Express $y = \left( \frac{1+x}{1+2x} \right)^{1/2}$ .
- Apply the chain rule and quotient rule to find $\frac{d y}{d x}$ .
- Simplify the result to get $\frac{d y}{d x} = \frac{x}{2 \sqrt{(1+x)(1+2x)^3}}$ .
For part ii:
- Express $y = \sin \left( \sqrt{\frac{1+2x}{1+x}} \right)$ .
- Apply the chain rule and quotient rule to find $\frac{d y}{d x}$ .
- Simplify the result to get $\frac{d y}{d x} = \frac{\cos \left( \sqrt{\frac{1+2x}{1+x}} \right)}{2 \sqrt{\frac{1+2x}{1+x}} (1+x)^{3/2}}$ .