Let f(x)=(sqrtx)/(sin(x)). Find f^(')(x). Choose 1 answer: (A) (1)/(2sqrtxcos(x)) (B) (cos(x))/(2sqrtx) (C) (sin(x)-2x cos(x))/(2sqrtxsin^(2)(x)) (D) (sin(x)+2x cos(x))/(2sqrtxsin^(2)(x))

Identify Function: Identify the function to differentiate. f(x) = (sqrtx)/(sin(x)) is a quotient of two functions, where the numerator is sqrtx and the denominator is sin(x).Apply Quotient Rule: Apply the quotient rule for differentiation. The quotient rule states that (u/v)' = (u'v - uv') / v^2, where u is the numerator and v is the denominator. Let u = sqrtx and v = sin(x). We need to find u' and v'.Differentiate u and v: Differentiate u and v. u = sqrtx = x^(1/2), so u' = (1/2)x^(-1/2) = 1/(2sqrtx). v = sin(x), so v' = cos(x).Apply Derivatives: Apply the derivatives to the quotient rule. f'(x) = (u'v - uv') / v^2 = (1/(2sqrtx) * sin(x) - sqrtx * cos(x)) / sin^2(x) = (sin(x)/(2sqrtx) - x * cos(x)/(sqrtx)) / sin^2(x) = (sin(x) - 2x * cos(x)) / (2sqrtx * sin^2(x))Choose Correct Answer: Choose the correct answer. The correct answer is (C) (sin(x) - 2x * cos(x)) / (2sqrtx * sin^2(x)).

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Let
f(x)=(sqrtx)/(sin(x)).
Find
f^(')(x).
Choose 1 answer:
(A)
(1)/(2sqrtxcos(x))
(B)
(cos(x))/(2sqrtx)
(C)
(sin(x)-2x cos(x))/(2sqrtxsin^(2)(x))
(D)
(sin(x)+2x cos(x))/(2sqrtxsin^(2)(x))

Let $f(x)=\frac{\sqrt{x}}{\sin (x)}$ . $\newline$ Find $f^{\prime}(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{2 \sqrt{x} \cos (x)}$ $\newline$ (B) $\frac{\cos (x)}{2 \sqrt{x}}$ $\newline$ (C) $\frac{\sin (x)-2 x \cos (x)}{2 \sqrt{x} \sin ^{2}(x)}$ $\newline$ (D) $\frac{\sin (x)+2 x \cos (x)}{2 \sqrt{x} \sin ^{2}(x)}$

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Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. Let

f(x)=\frac{\sqrt{x}}{\sin (x)}

\newline

Find

f^{\prime}(x)

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{2 \sqrt{x} \cos (x)}

\newline

(B)

\frac{\cos (x)}{2 \sqrt{x}}

\newline

(C)

\frac{\sin (x)-2 x \cos (x)}{2 \sqrt{x} \sin ^{2}(x)}

\newline

(D)

\frac{\sin (x)+2 x \cos (x)}{2 \sqrt{x} \sin ^{2}(x)}

Identify Function: Identify the function to differentiate. $f(x) = \frac{\sqrt{x}}{\sin(x)}$ is a quotient of two functions, where the numerator is $\sqrt{x}$ and the denominator is $\sin(x)$ .
Apply Quotient Rule: Apply the quotient rule for differentiation. $\newline$ The quotient rule states that $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$ , where $u$ is the numerator and $v$ is the denominator. $\newline$ Let $u = \sqrt{x}$ and $v = \sin(x)$ . We need to find $u'$ and $v'$ .
Differentiate $u$ and $v$ : Differentiate $u$ and $v$ .
$u = \sqrt{x} = x^{1/2}$ , so $u' = (1/2)x^{-1/2} = 1/(2\sqrt{x})$ .
$v = \sin(x)$ , so $v' = \cos(x)$ .
Apply Derivatives: Apply the derivatives to the quotient rule. $\newline$ $f'(x) = \frac{u'v - uv'}{v^2}$ $\newline$ = $\frac{\frac{1}{2\sqrt{x}} \cdot \sin(x) - \sqrt{x} \cdot \cos(x)}{\sin^2(x)}$ $\newline$ = $\frac{\frac{\sin(x)}{2\sqrt{x}} - \frac{x \cdot \cos(x)}{\sqrt{x}}}{\sin^2(x)}$ $\newline$ = $\frac{\sin(x) - 2x \cdot \cos(x)}{2\sqrt{x} \cdot \sin^2(x)}$
Choose Correct Answer: Choose the correct answer. $\newline$ The correct answer is (C) $(\sin(x) - 2x \cdot \cos(x)) / (2\sqrt{x} \cdot \sin^2(x))$ .