frac{d}{dx} left( sqrt{x} + frac{1}{sqrt{x}} | vedclass

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

Question

(A) Given expression: $y = \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2$ Expand the square using the identity $(a+b)^2 = a^2 + b^2 + 2ab$: $y = (\sq

Vedclass · Accepted Answer

$1 - \frac{1}{x^2}$

Vedclass · Answer

(A) Given expression: $y = \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2$ Expand the square using the identity $(a+b)^2 = a^2 + b^2 + 2ab$: $y = (\sqrt{x})^2 + \left( \frac{1}{\sqrt{x}} \right)^2 + 2(\sqrt{x}) \left( \frac{1}{\sqrt{x}} \right)$ $y = x + \frac{1}{x} + 2$ Now,differentiate with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1}) + \frac{d}{dx}(2)$ $\frac{dy}{dx} = 1 - x^{-2} + 0$ $\frac{dy}{dx} = 1 - \frac{1}{x^2}$ Thus,the correct option is $A$.

$\frac{d}{dx} \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 = $

Similar Questions

If $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}}$,then $f'(a) = $

Let $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$,where $x \in R$. Then $f'(10)$ is equal to:

If $f(x) = (\log _{\cot x}\tan x)(\log _{\tan x}\cot x)^{-1}$,then $f'(2) = $

If $y = \sin^{98}(x) \cdot \cos^{39}(x)$,then find $\frac{dy}{dx}$.

Suppose that $h(x) = f(x) \cdot g(x)$ and $F(x) = f(g(x))$,where $f(2) = 3$,$g(2) = 5$,$g'(2) = 4$,$f'(2) = -2$,and $f'(5) = 11$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module