Let h(x)=sqrtxln(x). Find h^(')(x). Choose 1 answer: (A) (ln(x))/(2sqrtx)+(1)/(sqrtx) (B) (1)/(2sqrtx)+(1)/(x) (C) (1)/(2sqrtx)*(1)/(x) (D) (sqrtxln(x))/(2)+(1)/(sqrtx)

Define Functions: The product rule states that (fg)' = f'g + fg'. Let f(x) = sqrt(x) and g(x) = ln(x). We need to find f'(x) and g'(x).Find f'(x): Differentiate f(x) = sqrt(x). We can write sqrt(x) as x^(1/2) and use the power rule. f'(x) = (1/2)x^(-1/2).Find g'(x): Differentiate g(x) = ln(x). The derivative of ln(x) is 1/x. So, g'(x) = 1/x.Apply Product Rule: Now apply the product rule: h'(x) = f'(x)g(x) + f(x)g'(x). Plug in the derivatives we found: h'(x) = (1/2)x^(-1/2)ln(x) + x^(1/2)(1/x).Simplify Expression: Simplify the expression: h'(x) = (1/2)x^(-1/2)ln(x) + (1/2)x^(-1/2). This is because x^(1/2) * (1/x) = x^(-1/2).

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Let
h(x)=sqrtxln(x).
Find
h^(')(x).
Choose 1 answer:
(A)
(ln(x))/(2sqrtx)+(1)/(sqrtx)
(B)
(1)/(2sqrtx)+(1)/(x)
(C)
(1)/(2sqrtx)*(1)/(x)
(D)
(sqrtxln(x))/(2)+(1)/(sqrtx)

Let $h(x)=\sqrt{x} \ln (x)$ . $\newline$ Find $h^{\prime}(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{\ln (x)}{2 \sqrt{x}}+\frac{1}{\sqrt{x}}$ $\newline$ (B) $\frac{1}{2 \sqrt{x}}+\frac{1}{x}$ $\newline$ (C) $\frac{1}{2 \sqrt{x}} \cdot \frac{1}{x}$ $\newline$ (D) $\frac{\sqrt{x} \ln (x)}{2}+\frac{1}{\sqrt{x}}$

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Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. Let

h(x)=\sqrt{x} \ln (x)

\newline

Find

h^{\prime}(x)

\newline

Choose

1

answer:

\newline

(A)

\frac{\ln (x)}{2 \sqrt{x}}+\frac{1}{\sqrt{x}}

\newline

(B)

\frac{1}{2 \sqrt{x}}+\frac{1}{x}

\newline

(C)

\frac{1}{2 \sqrt{x}} \cdot \frac{1}{x}

\newline

(D)

\frac{\sqrt{x} \ln (x)}{2}+\frac{1}{\sqrt{x}}

Define Functions: The product rule states that $(fg)' = f'g + fg'$ . Let $f(x) = \sqrt{x}$ and $g(x) = \ln(x)$ . We need to find $f'(x)$ and $g'(x)$ .
Find $f'(x)$ : Differentiate $f(x) = \sqrt{x}$ . We can write $\sqrt{x}$ as $x^{(1/2)}$ and use the power rule. $f'(x) = (\frac{1}{2})x^{(-1/2)}$ .
Find $g'(x)$ : Differentiate $g(x) = \ln(x)$ . The derivative of $\ln(x)$ is $\frac{1}{x}$ . So, $g'(x) = \frac{1}{x}$ .
Apply Product Rule: Now apply the product rule: $h'(x) = f'(x)g(x) + f(x)g'(x)$ . Plug in the derivatives we found: $h'(x) = (\frac{1}{2})x^{(-\frac{1}{2})}\ln(x) + x^{(\frac{1}{2})}(\frac{1}{x})$ .
Simplify Expression: Simplify the expression: $h'(x) = (\frac{1}{2})x^{(-\frac{1}{2})}\ln(x) + (\frac{1}{2})x^{(-\frac{1}{2})}$ . This is because $x^{(\frac{1}{2})} \cdot (\frac{1}{x}) = x^{(-\frac{1}{2})}$ .