x = frac{1-sqrt{y}}{1+sqrt{y}} Rightarrow fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given,$x = \frac{1-\sqrt{y}}{1+\sqrt{y}}$.Applying componendo and dividendo,we get:$\frac{1+x}{1-x} = \frac{(1+\sqrt{y})+(1-\sqrt{y})}{(1+\sqrt{y}

Vedclass · Accepted Answer

$\frac{4(x-1)}{(1+x)^3}$

Vedclass · Answer

(B) Given,$x = \frac{1-\sqrt{y}}{1+\sqrt{y}}$.Applying componendo and dividendo,we get:$\frac{1+x}{1-x} = \frac{(1+\sqrt{y})+(1-\sqrt{y})}{(1+\sqrt{y})-(1-\sqrt{y})} = \frac{2}{2\sqrt{y}} = \frac{1}{\sqrt{y}}$.Squaring both sides,we get $\sqrt{y} = \frac{1-x}{1+x}$,so $y = \left(\frac{1-x}{1+x}\right)^2$.Now,differentiating $y$ with respect to $x$ using the chain rule:$\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$.Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:$\frac{d}{dx}\left(\frac{1-x}{1+x}\right) = \frac{(-1)(1+x) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = \frac{-2}{(1+x)^2}$.Therefore,$\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \left(\frac{-2}{(1+x)^2}\right) = \frac{-4(1-x)}{(1+x)^3} = \frac{4(x-1)}{(1+x)^3}$.

List $I$	List $II$
$(A)$ Fleming's left hand rule	$(i)$ Direction of induced current
$(B)$ Right hand thumb rule	(ii) Magnitude and direction of magnetic induction
$(C)$ Biot-Savart law	(iii) Direction of force due to magnetic induction
$(D)$ Fleming's right hand rule	(iv) Direction of magnetic lines due to current

$x = \frac{1-\sqrt{y}}{1+\sqrt{y}} \Rightarrow \frac{dy}{dx}$ is equal to

Similar Questions

Water is conveyed through a uniform tube of $8 \text{ cm}$ in diameter and $3140 \text{ m}$ in length at the rate of $2 \times 10^{-3} \text{ m}^3/\text{s}$. The pressure required to maintain the flow is (Viscosity of water $= 10^{-3} \text{ SI units}$):

In the given circuit,the readings of voltmeters $V_1$ and $V_2$ are $300 \, V$ each. The readings of the voltmeter $V_3$ and ammeter $A$ are respectively:

If $f(x) = \begin{cases} \frac{x-1}{2x^2-7x+5}, & \text{for } x \neq 1 \\ -\frac{1}{3}, & \text{for } x=1 \end{cases}$, then $f'(1)$ is equal to:

The part of the fallopian tube closest to the ovary is -

Vedclass Test Series

Exam Paper Generator

Online Exam Module