Question

y = √x + 1/√x হলে প্রমাণ করাে যে, 2x * dy/dx + y = 2√x​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\sf{y=\sqrt{x}+\dfrac{1}{\sqrt{x}}}$

$\sf{\implies\,y^2=\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2}$

$\sf{\implies\,y^2=x+\dfrac{1}{x}+2}$

$\sf{\implies\,2y\,\dfrac{dy}{dx}=1-\dfrac{1}{x^2}}$

$\sf{\implies\,2\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)\,\dfrac{dy}{dx}=\dfrac{x^2-1}{x^2}}$

$\sf{\implies\,2\left(\dfrac{x+1}{\sqrt{x}}\right)\,\dfrac{dy}{dx}=\dfrac{(x+1)(x-1)}{x^2}}$

$\sf{\implies\,\dfrac{2}{\sqrt{x}}\,\dfrac{dy}{dx}=\dfrac{x-1}{x^2}}$

$\sf{\implies\,\dfrac{2\,x^2}{\sqrt{x}}\,\dfrac{dy}{dx}=x-1}$

$\sf{\implies\,2\,x\sqrt{x}\,\dfrac{dy}{dx}=x-1}$

$\sf{\implies\,2x\,\dfrac{dy}{dx}=\dfrac{x-1}{\sqrt{x}}}$

$\sf{\implies\,2x\,\dfrac{dy}{dx}=\sqrt{x}-\dfrac{1}{\sqrt{x}}}$

$\sf{\implies\,2x\,\dfrac{dy}{dx}=2\sqrt{x}-\sqrt{x}-\dfrac{1}{\sqrt{x}}}$

$\sf{\implies\,2x\,\dfrac{dy}{dx}=2\sqrt{x}-\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)}$

$\sf{\implies\,2x\,\dfrac{dy}{dx}=2\sqrt{x}-y}$

$\sf{\implies\,2x\,\dfrac{dy}{dx}+y=2\sqrt{x}}$