How do you differentiate #f(x)= 5ln(sinx/x) #?

Question

How do you differentiate #f(x)= 5ln(sinx/x)  #?

Paisley Johnson · Accepted Answer

The derivative is #f'(x) = (xcosx - sinx)/(xsinx)#

The derivative of #sinx/x # is taken using quotient rule and multiplied with the derivative of #lnu# where #u= sinx/x# which is #1/u#.

Hope that helps, Cheers.

Ezekiel Alexander · Answer

undefined To differentiate $ f(x) = 5 \ln\left(\frac{\sin x}{x}\right) $, use the chain rule and the derivative of the natural logarithm. The derivative is $ f'(x) = 5 \cdot \frac{1}{\frac{\sin x}{x}} \cdot \left(\frac{\cos x \cdot x - \sin x \cdot 1}{x^2}\right) $. Simplify this to get $ f'(x) = 5 \cdot \frac{x \cdot \cos x - \sin x}{x \cdot \sin x} $.

Paisley Johnson · Answer

undefined To differentiate $ f(x) = 5\ln\left(\frac{\sin x}{x}\right) $, we will use the properties of logarithmic differentiation and the chain rule.

Given:

$$ f(x) = 5\ln\left(\frac{\sin x}{x}\right) $$

We can rewrite $ f(x) $ as:

$$ f(x) = 5\ln(\sin x) - 5\ln(x) $$

Now, differentiate each term separately:

1. Differentiate $ 5\ln(\sin x) $:
$$ \frac{d}{dx} (5\ln(\sin x)) = 5 \cdot \frac{1}{\sin x} \cdot \cos x $$

2. Differentiate $ -5\ln(x) $:
$$ \frac{d}{dx} (-5\ln(x)) = -\frac{5}{x} $$

Combining these results, we get the derivative of $ f(x) $ as:

$$ f'(x) = 5 \cdot \frac{\cos x}{\sin x} - \frac{5}{x} $$

$$ f'(x) = \frac{5\cos x}{\sin x} - \frac{5}{x} $$

$$ f'(x) = 5\cot x - \frac{5}{x} $$

So, the derivative of $ f(x) = 5\ln\left(\frac{\sin x}{x}\right) $ is $ f'(x) = 5\cot x - \frac{5}{x} $.