If y = e^{sqrt{x}} + e^{-sqrt{x}},then frac{d | vedclass

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

Question

(D) Given $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{d}{dx}(e^{\sqrt{x}}) + \f

Vedclass · Accepted Answer

$A$ and $C$

Vedclass · Answer

(D) Given $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{d}{dx}(e^{\sqrt{x}}) + \frac{d}{dx}(e^{-\sqrt{x}})$$\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{d}{dx}(\sqrt{x}) + e^{-\sqrt{x}} \cdot \frac{d}{dx}(-\sqrt{x})$$\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} + e^{-\sqrt{x}} \cdot (-\frac{1}{2\sqrt{x}})$$\frac{dy}{dx} = \frac{e^{\sqrt{x}} - e^{-\sqrt{x}}}{2\sqrt{x}}$ (This matches option $A$).Now,consider $y^2 - 4 = (e^{\sqrt{x}} + e^{-\sqrt{x}})^2 - 4$$y^2 - 4 = e^{2\sqrt{x}} + e^{-2\sqrt{x}} + 2 - 4 = e^{2\sqrt{x}} + e^{-2\sqrt{x}} - 2$$y^2 - 4 = (e^{\sqrt{x}} - e^{-\sqrt{x}})^2$Taking the square root,$\sqrt{y^2 - 4} = e^{\sqrt{x}} - e^{-\sqrt{x}}$.Substituting this into the derivative expression:$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \sqrt{y^2 - 4}$ (This matches option $C$).Since both $A$ and $C$ are correct,the correct option is $D$.

If $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$,then $\frac{dy}{dx}$ equals:

Similar Questions

If $f$ and $g$ are differentiable functions satisfying $g^{\prime}(a)=2$,$g(a)=b$,and $f \circ g = I$,where $I$ is an identity function,then $f^{\prime}(b)$ is equal to

If $y = \sin^{-1}\sqrt{1 - x} + \cos^{-1}\sqrt{x}$,then $\frac{dy}{dx} = $

If $f^{\prime}(x) = \tan^{-1}(\sec x + \tan x)$,$\frac{-\pi}{2} < x < \frac{\pi}{2}$ and $f(0) = 0$,then $f(1) =$

If $y = \tan^{-1}(\frac{1}{x^2+x+1}) + \tan^{-1}(\frac{1}{x^2+3x+3}) + \tan^{-1}(\frac{1}{x^2+5x+7}) + \dots$ upto $n$ terms, then $y'(0) = $

If $y = (\cos x^{2})^{2}$,then $\frac{dy}{dx}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module