Question

please look at the attachment and give me answer with steps in a basic form I've just got admitted in 11th std​

if you are army then answer me

Answer 1

Answer:

$\frac{dy}{dx} = \frac{A}{x^{2} } -\frac{1}{2x\sqrt{x} }$

Explanation:

Given,

$y = \frac{\sqrt{x}-A}{x}$ , which can be further simplified as

$y = \frac{\sqrt{x}}{x} - \frac{A}{x} =\frac{1}{\sqrt{x}} - \frac{A}{x}$

$y=x^{-\frac{1}{2} } - Ax^{-1}$

[Remember,$\sqrt{x}$ can be represented as $x^{\frac{1}{2} }$ , and also $x^{-n} = \frac{1}{x^{n} }$

On differentiating it w.r.t 'dx',

$\frac{d(y)}{dx}= \frac{d}{dx} (x^{-\frac{1}{2} } - Ax^{-1})$

Applying the formula, $\frac{d(x^{n} )}{dx} = nx^{n-1}$, we can solve it as follows:

$\frac{d(y)}{dx} = -\frac{1}{2} x^{\frac{-1}{2}- 1 } - (-1)(A)(x^{-1-1)}$

$\frac{d(y)}{dx} = \frac{-1}{2} x^{\frac{-3}{2} } - (-Ax^{-2})$

$\frac{d(y)}{dx} = \frac{-1}{2x^{3/2} } + \frac{A}{x^{2} }$

$\frac{d(y)}{dx} = \frac{-1}{2\sqrt{x^{3} } } + \frac{A}{x^{2} }$

$\frac{d(y)}{dx} = \frac{-1 }{2x\sqrt{x} } + \frac{A}{x^{2} }$

$\frac{d(y)}{dx} = \frac{-x^{2} + 2x\sqrt{x} A}{2x^{3}\sqrt{x} }$

$\frac{d(y)}{dx} = \frac{x (2A\sqrt{x} - x)}{2x^{3}\sqrt{x} }$

$\frac{dy}{dx} = \frac{2A\sqrt{x} - x}{2x^{2} \sqrt{x} }$

Further, on splitting the numerator, we get

$\frac{dy}{dx} = \frac{A}{x^{2} } -\frac{1}{2x\sqrt{x} }$

Hope this helps!