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

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

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Full solution

Q. Let

y=\sqrt{x} \cos (x)

.

\newline

Find

\frac{d y}{d x}

.

\newline

Choose

1

answer:

\newline

(A)

-\frac{1}{\sqrt{x}}+\sin (x)

\newline

(B)

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

\newline

(C)

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

\newline

(D)

-\frac{\sin (x)}{\sqrt{x}}

Apply Product Rule: Use the product rule for differentiation, which states that $(fg)' = f'g + fg'$ , where $f$ and $g$ are functions of $x$ . Here, $f(x) = \sqrt{x}$ and $g(x) = \cos(x)$ .
Differentiate $\sqrt{x}$ : Differentiate $f(x) = \sqrt{x}$ with respect to $x$ to get $f'(x)$ . The derivative of $\sqrt{x}$ is $(\frac{1}{2})x^{(-\frac{1}{2})}$ .
Differentiate $\cos(x)$ : Differentiate $g(x) = \cos(x)$ with respect to $x$ to get $g'(x)$ . The derivative of $\cos(x)$ is $-\sin(x)$ .
Apply Product Rule: Apply the product rule: $(\sqrt{x}\cos(x))' = \frac{1}{2}x^{-\frac{1}{2}}\cos(x) + \sqrt{x}(-\sin(x))$ .
Simplify Expression: Simplify the expression: (\frac{\(1\)}{\(2\)})x^{(-\frac{\(1\)}{\(2\)})}\cos(x) - x^{(\frac{\(1\)}{\(2\)})}\sin(x)\.
Write in Terms of \(\sqrt{x}: Write the expression in terms of $\sqrt{x}$ : $\frac{\cos(x)}{2\sqrt{x}} - \sqrt{x}\sin(x)$ .
Match with Answer Choices: Match the simplified derivative with the given answer choices.

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