Question

If y = 2√x÷ 1–x', show that dy÷dx = x+1 ÷ √x(1–x)²​

Answer 1

Answer:  see proof below

Step-by-step explanation:

Use the Quotient rule for derivatives:

$\text{If}\ y=\dfrac{a}{b}\quad \text{then}\ y'=\dfrac{a'b-ab'}{b^2}$

Given: $y=\dfrac{2\sqrtx}{1-x}$

$\sqrtx$$a=2\sqrt x\qquad \rightarrow \qquad a'=\dfrac{1}{\sqrt x}\\\\b=1-x\qquad \rightarrow \qquad b'=-1$        

$y'=\dfrac{\dfrac{1-x}{\sqrt x}-(-2\sqrt x)}{(1-x)^2}\\\\\\.\quad =\dfrac{\dfrac{1-x}{\sqrt x}-(-2\sqrt x)\bigg(\dfrac{\sqrt x}{\sqrt x}\bigg)}{(1-x)^2}\\\\\\.\quad =\dfrac{1-x+2x}{\sqrt x(1-x)^2}\\\\\\.\quad =\dfrac{x+1}{\sqrt x(1-x)^2}$

LHS = RHS:  $\dfrac{x+1}{\sqrt x(1-x)^2}=\dfrac{x+1}{\sqrt x(1-x)^2}\qquad \checkmark$