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Use logarithmic differentiation to find the derivative of
y with respect to the independent variable.
qquad

y=x^(ln x)
A)
x^(ln x-1)ln x
B)
2x^(ln x-1)ln x
C)
(2ln x)/(x)
D)
(ln x)^(2)

Use logarithmic differentiation to find the derivative of $y$ with respect to the independent variable. $\qquad$ $\newline$ $y=x^{\ln x}$ $\newline$ A) $x^{\ln x-1} \ln x$ $\newline$ B) $2 x^{\ln x-1} \ln x$ $\newline$ C) $\frac{2 \ln x}{x}$ $\newline$ D) $(\ln x)^{2}$

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Sum of finite series not start from 1

Full solution

Q. Use logarithmic differentiation to find the derivative of

y

with respect to the independent variable.

\qquad

\newline

y=x^{\ln x}

\newline

A)

x^{\ln x-1} \ln x

\newline

B)

2 x^{\ln x-1} \ln x

\newline

C)

\frac{2 \ln x}{x}

\newline

D)

(\ln x)^{2}

Apply logarithmic differentiation: Step $1$ : Apply logarithmic differentiation. $\newline$ Take the natural logarithm of both sides: $\newline$ $\ln(y) = \ln(x^{\ln x})$ $\newline$ Using the power rule for logarithms, simplify: $\newline$ $\ln(y) = (\ln x)(\ln x) = (\ln x)^2$
Differentiate with respect to $x$ : Step $2$ : Differentiate both sides with respect to $x$ . Using the chain rule on the left side: $\frac{d}{dx}[\ln(y)] = \frac{1}{y} \cdot \frac{dy}{dx}$ Differentiate the right side using the product rule: $\frac{d}{dx}[(\ln x)^2] = 2(\ln x) \cdot \frac{d}{dx}[\ln x]$ Since $\frac{d}{dx}[\ln x] = \frac{1}{x}$ , This becomes $\frac{2(\ln x)}{x}$
Solve for $\frac{dy}{dx}$ : Step $3$ : Solve for $\frac{dy}{dx}$ . $\newline$ From the differentiation: $\newline$ $\frac{1}{y} \cdot \frac{dy}{dx} = \frac{2(\ln x)}{x}$ $\newline$ Multiply both sides by $y$ to solve for $\frac{dy}{dx}$ : $\newline$ $\frac{dy}{dx} = y \cdot \left(\frac{2(\ln x)}{x}\right)$ $\newline$ Substitute back for $y = x^{(\ln x)}$ : $\newline$ $\frac{dy}{dx} = x^{(\ln x)} \cdot \left(\frac{2(\ln x)}{x}\right)$ $\newline$ Simplify: $\newline$ $\frac{dy}{dx} = 2x^{(\ln x - 1)}(\ln x)$

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