If y = e^{sqrt{x}},then frac{dy}{dx} equals: | vedclass

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

Question

(A) Given the function $y = e^{\sqrt{x}}$.To find the derivative $\frac{dy}{dx}$,we apply the chain rule.Let $u = \sqrt{x} = x^{1/2}$. Then $y = e^u$.

Vedclass · Accepted Answer

$\frac{e^{\sqrt{x}}}{2\sqrt{x}}$

Vedclass · Answer

(A) Given the function $y = e^{\sqrt{x}}$.To find the derivative $\frac{dy}{dx}$,we apply the chain rule.Let $u = \sqrt{x} = x^{1/2}$. Then $y = e^u$.Using the chain rule: $\frac{dy}{dx} = \frac{dy}{du} 	imes \frac{du}{dx}$.$\frac{dy}{du} = \frac{d}{du}(e^u) = e^u = e^{\sqrt{x}}$.$\frac{du}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.Therefore,$\frac{dy}{dx} = e^{\sqrt{x}} 	imes \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}$.

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

Similar Questions

If $f(x)=x^{n}$, where $n$ is a non-negative integer, then the values of $n$ for which $f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)$ for all $\alpha, \beta > 0$ is:

The derivative of ${x^6} + {6^x}$ with respect to $x$ is:

If $y = \frac{(1 - x)^2}{x^2}$,then $\frac{dy}{dx}$ is

If $y = \frac{ax+b}{cx+d}$ and $\frac{dx}{dy} = \frac{ad-bc}{Py^2+Qy+R}$,then $P+Q+R =$

If $G(x) = -\sqrt{25 - x^2}$,then $\mathop{\lim}\limits_{x \to 1} \frac{G(x) - G(1)}{x - 1} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module