Question

y=√1-2x/1+2x find dy/dx​

Answer 1

Step-by-step explanation:

y=(2x+1)^1/3

dy/dx = 1/3 (2x+1)^1/3

=1/3 (2x+1)^-2/3

= 1/3(2x+1)^2/3

=1/3√(2x+1)^2

Answer 2

The Derivative of $y = \frac{\sqrt{1-2x}}{2x+1}$ is  $\frac{dy}{dx} = \frac{2x-3}{\sqrt{1-2x}\left(2x+1\right)^2}$

Step-by-step explanation:

Given: $y = \frac{\sqrt{1-2x}}{2x+1}$

To Find: Derivative $\frac{dy}{dx}$

Solution:

• Finding derivative of $\frac{\sqrt{1-2x}}{2x+1}$

We have the function, $y = \dfrac{\sqrt{1-2x}}{2x+1}$ such that the derivative will be as follows,

$\dfrac{dy}{dx} = \dfrac{d}{dx} \Big(\dfrac{\sqrt{1-2x}}{2x+1} \Big)$

Applying the quotient rule of differentiation, we get

$\Rightarrow \dfrac{dy}{dx} = \dfrac{\Big(\dfrac{d}{dx} (\sqrt{1-2x} ) \Big) \cdot (2x+1) - (\sqrt{1-2x} ) \cdot \Big(\dfrac{d}{dx} (2x+1) \Big)}{(2x+1)^2}$

$\Rightarrow \dfrac{dy}{dx} = \dfrac{\dfrac{1}{2} ({1-2x} )^{\frac{1}{2}-1 } \cdot \Big(\dfrac{d}{dx} ({1-2x} ) \Big) \cdot (2x+1) - (\sqrt{1-2x} ) \cdot \Big(2\dfrac{d(x)}{dx} + \dfrac{d(1)}{dx} \Big)}{(2x+1)^2}$

Solving it further, we get

$\Rightarrow \dfrac{dy}{dx} = \dfrac{-\frac{2x+1}{\sqrt{1-2x}}-2\sqrt{1-2x}}{\left(2x+1\right)^2}$

Simplifying and calculating the above expression, we get

$\Rightarrow \dfrac{dy}{dx} = \dfrac{2x-3}{\sqrt{1-2x}\left(2x+1\right)^2}$

Hence, the Derivative of $y = \frac{\sqrt{1-2x}}{2x+1}$ is  $\frac{dy}{dx} = \frac{2x-3}{\sqrt{1-2x}\left(2x+1\right)^2}$