Find (d)/(dx)((e^(x))/(sqrtx)). Choose 1 answer: (A) e^(x)-(1)/(2sqrtx) (B) (e^(x)(sqrtx-(1)/(2sqrtx)))/(x) (C) 2sqrtxe^(x) (D) e^(x)(sqrtx-(1)/(2sqrtx))

Apply Quotient Rule: Apply the quotient rule for differentiation, which states that the derivative of a function h(x) = f(x)/g(x) is given by h'(x) = (g(x)f'(x) - f(x)g'(x))/(g(x))^2. Let f(x) = e^x and g(x) = sqrt(x), so we need to find f'(x) and g'(x).Find f'(x): Find the derivative of f(x) = e^x with respect to x. The derivative of e^x with respect to x is e^x. So, f'(x) = e^x.Find g'(x): Find the derivative of g(x) = sqrt(x) with respect to x. The derivative of sqrt(x) is (1/2)x^(-1/2). So, g'(x) = (1/2)x^(-1/2) = 1/(2sqrt(x)).Apply Quotient Rule: Apply the quotient rule using the derivatives from steps 2 and 3. h'(x) = (g(x)f'(x) - f(x)g'(x))/(g(x))^2 h'(x) = (sqrt(x)e^x - e^x(1/(2sqrt(x))))/(sqrt(x))^2Simplify Expression: Simplify the expression. h'(x) = (sqrt(x)e^x - e^x/(2sqrt(x)))/(x) h'(x) = (e^x(sqrt(x) - 1/(2sqrt(x))))/x h'(x) = e^x(sqrt(x) - 1/(2sqrt(x)))/xCheck Answer Choices: Check the answer choices to see which one matches our result. The correct answer is (D) e^(x)(sqrtx-(1)/(2sqrtx)).

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Find
(d)/(dx)((e^(x))/(sqrtx)).
Choose 1 answer:
(A)
e^(x)-(1)/(2sqrtx)
(B)
(e^(x)(sqrtx-(1)/(2sqrtx)))/(x)
(C)
2sqrtxe^(x)
(D)
e^(x)(sqrtx-(1)/(2sqrtx))

Find $\frac{d}{d x}\left(\frac{e^{x}}{\sqrt{x}}\right)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $e^{x}-\frac{1}{2 \sqrt{x}}$ $\newline$ (B) $\frac{e^{x}\left(\sqrt{x}-\frac{1}{2 \sqrt{x}}\right)}{x}$ $\newline$ (C) $2 \sqrt{x} e^{x}$ $\newline$ (D) $e^{x}\left(\sqrt{x}-\frac{1}{2 \sqrt{x}}\right)$

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Math Problems

Calculus

Find derivatives of using multiple formulae

Full solution

Q. Find

\frac{d}{d x}\left(\frac{e^{x}}{\sqrt{x}}\right)

\newline

Choose

1

answer:

\newline

(A)

e^{x}-\frac{1}{2 \sqrt{x}}

\newline

(B)

\frac{e^{x}\left(\sqrt{x}-\frac{1}{2 \sqrt{x}}\right)}{x}

\newline

(C)

2 \sqrt{x} e^{x}

\newline

(D)

e^{x}\left(\sqrt{x}-\frac{1}{2 \sqrt{x}}\right)

Apply Quotient Rule: Apply the quotient rule for differentiation, which states that the derivative of a function $h(x) = \frac{f(x)}{g(x)}$ is given by $h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$ . Let $f(x) = e^x$ and $g(x) = \sqrt{x}$ , so we need to find $f'(x)$ and $g'(x)$ .
Find $f'(x)$ : Find the derivative of $f(x) = e^x$ with respect to $x$ . The derivative of $e^x$ with respect to $x$ is $e^x$ . So, $f'(x) = e^x$ .
Find $g'(x)$ : Find the derivative of $g(x) = \sqrt{x}$ with respect to $x$ . The derivative of $\sqrt{x}$ is $(1/2)x^{(-1/2)}$ . So, $g'(x) = (1/2)x^{(-1/2)} = 1/(2\sqrt{x})$ .
Apply Quotient Rule: Apply the quotient rule using the derivatives from steps $2$ and $3$ . $\newline$ $h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$ $\newline$ $h'(x) = \frac{\sqrt{x}e^x - e^x(\frac{1}{2\sqrt{x}})}{(\sqrt{x})^2}$
Simplify Expression: Simplify the expression. $\newline$ $h'(x) = \frac{\sqrt{x}e^x - \frac{e^x}{2\sqrt{x}}}{x}$ $\newline$ $h'(x) = \frac{e^x(\sqrt{x} - \frac{1}{2\sqrt{x}})}{x}$ $\newline$ $h'(x) = \frac{e^x(\sqrt{x} - \frac{1}{2\sqrt{x}})}{x}$
Check Answer Choices: Check the answer choices to see which one matches our result. $\newline$ The correct answer is (D) $e^{x}(\sqrt{x}-\frac{1}{2\sqrt{x}})$ .