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A curve $C$ has equation $\newline$ $y=x^{(\sin x)}\quad x > 0\quad y > 0$ $\newline$ (a) Find, by firstly taking natural logarithms, an expression for $(dy)/(dx)$ in terms of $x$ and $y$ .

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Complex conjugate theorem

Full solution

Q. A curve

C

has equation

\newline

y=x^{(\sin x)}\quad x > 0\quad y > 0

\newline

(a) Find, by firstly taking natural logarithms, an expression for

(dy)/(dx)

in terms of

x

and

y

.

Apply Logarithm Property: To find the derivative of $y$ with respect to $x$ , we will first take the natural logarithm of both sides of the equation $y = x^{(\sin x)}$ to make use of the properties of logarithms. $\newline$ We apply the logarithm to both sides: $\newline$ $\ln(y) = \ln(x^{(\sin x)})$ $\newline$ Using the property of logarithms that allows us to bring the exponent down in front of the log, we get: $\newline$ $\ln(y) = \sin(x) \cdot \ln(x)$
Differentiate with Chain and Product Rule: Now we differentiate both sides of the equation with respect to $x$ . The left side will use the chain rule, and the right side will use the product rule. $\newline$ Differentiating the left side: $\newline$ $\frac{d}{dx} [\ln(y)] = \frac{1}{y} \cdot \frac{dy}{dx}$ $\newline$ Differentiating the right side: $\newline$ \frac{d}{dx} [\sin(x) \cdot \ln(x)] = \cos(x) \cdot \ln(x) + \sin(x) \cdot \left(\frac{\(1\)}{x}\right)
Solve for \(\frac{dy}{dx}: Now we equate the derivatives from both sides to solve for $\frac{dy}{dx}$ . $\frac{1}{y} \cdot \frac{dy}{dx} = \cos(x) \cdot \ln(x) + \sin(x) \cdot \left(\frac{1}{x}\right)$ To isolate $\frac{dy}{dx}$ , we multiply both sides by $y$ : $\frac{dy}{dx} = y \cdot (\cos(x) \cdot \ln(x) + \sin(x) \cdot \left(\frac{1}{x}\right))$
Substitute back into Expression: We know from the original equation that $y = x^{(\sin x)}$ , so we can substitute this back into our expression for $\frac{dy}{dx}$ . $\frac{dy}{dx} = x^{(\sin x)} \cdot (\cos(x) \cdot \ln(x) + \sin(x) \cdot (\frac{1}{x}))$ This is the expression for the derivative of $y$ with respect to $x$ in terms of $x$ and $y$ .

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