Answered

Use logarithmic differentiation to find [tex]\frac{dy}{dx}[/tex] if [tex]y=\frac{\sin x}{x^3 \ln x}[/tex].



Answer :

To find [tex]\(\frac{dy}{dx}\)[/tex] for [tex]\(y = \frac{\sin x}{x^3 \ln x}\)[/tex] using logarithmic differentiation, follow these steps:

1. Write the original function:
[tex]\[ y = \frac{\sin x}{x^3 \ln x} \][/tex]

2. Take the natural logarithm of both sides:
[tex]\[ \ln y = \ln \left( \frac{\sin x}{x^3 \ln x} \right) \][/tex]

3. Use the properties of logarithms to simplify:
[tex]\[ \ln y = \ln (\sin x) - \ln (x^3) - \ln (\ln x) \][/tex]
Using the property [tex]\(\ln(a/b) = \ln a - \ln b\)[/tex], and [tex]\(\ln(a \cdot b) = \ln a + \ln b\)[/tex].

Since [tex]\(\ln (x^3) = 3 \ln x\)[/tex]:
[tex]\[ \ln y = \ln (\sin x) - 3 \ln x - \ln (\ln x) \][/tex]

4. Differentiate both sides with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{d}{dx} (\ln y) = \frac{d}{dx} \left( \ln (\sin x) - 3 \ln x - \ln (\ln x) \right) \][/tex]

On the left side, using the chain rule:
[tex]\[ \frac{d}{dx} (\ln y) = \frac{1}{y} \frac{dy}{dx} \][/tex]

On the right side:
[tex]\[ \frac{d}{dx} \left( \ln (\sin x) \right) = \frac{1}{\sin x} \cdot \cos x = \cot x \][/tex]
[tex]\[ \frac{d}{dx} \left( -3 \ln x \right) = -3 \cdot \frac{1}{x} = -\frac{3}{x} \][/tex]
[tex]\[ \frac{d}{dx} \left( -\ln (\ln x) \right) = -\frac{1}{\ln x} \cdot \frac{1}{x} = -\frac{1}{x \ln x} \][/tex]

Combining these, the right side becomes:
[tex]\[ \frac{dy}{dx} \left( \frac{1}{y} \right) = \cot x - \frac{3}{x} - \frac{1}{x \ln x} \][/tex]

5. Multiply both sides by [tex]\(y\)[/tex] to isolate [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \frac{dy}{dx} = y \left( \cot x - \frac{3}{x} - \frac{1}{x \ln x} \right) \][/tex]

6. Substitute [tex]\(y\)[/tex] back in terms of [tex]\(x\)[/tex]:
Recall that [tex]\(y = \frac{\sin x}{x^3 \ln x}\)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{\sin x}{x^3 \ln x} \left( \cot x - \frac{3}{x} - \frac{1}{x \ln x} \right) \][/tex]

Simplify this expression:
Note that [tex]\(\cot x = \frac{\cos x}{\sin x}\)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{\sin x}{x^3 \ln x} \left( \frac{\cos x}{\sin x} - \frac{3}{x} - \frac{1}{x \ln x} \right) \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{\sin x}{x^3 \ln x} \cdot \frac{\cos x}{\sin x} - \frac{\sin x}{x^3 \ln x} \cdot \frac{3}{x} - \frac{\sin x}{x^3 \ln x} \cdot \frac{1}{x \ln x} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{\cos x}{x^3 \ln x} - \frac{3 \sin x}{x^4 \ln x} - \frac{\sin x}{x^4 \ln^2 x} \][/tex]

Thus, the derivative is:
[tex]\[ \frac{dy}{dx} = \frac{\cos x}{x^3 \ln x} - \frac{3 \sin x}{x^4 \ln x} - \frac{\sin x}{x^4 \ln^2 x} \][/tex]

So, the final answer is:
[tex]\[ \frac{dy}{dx} = \frac{\cos x}{x^3 \ln x} - \frac{3 \sin x}{x^4 \ln x} - \frac{\sin x}{x^4 \ln^2 x}. \][/tex]