How do you differentiate #f(x)= ln(sin(x^2)/x) #?

Question

Emilia Aguilar · Accepted Answer

To differentiate the #ln#, we'll need quotient rule.
To differentiate #sin(x^2)#, we'll need chain rule as well.
To differentiate #sin(x^2)/x#, we'll need quotient rule.

We can rename #u=sin(x^2)/x# so that #f(x)=ln(u)#, which is differentiable. Also, we can rename #v=x^2# so we can differentiate #sin(v)# applying chain rule, as well. #(dy)/(dx)=1/u*(2xcos(x^2)*x-sin(x^2)*1)/x^2# #(dy)/(dx)=(2x^2cos(x^2)-sin(x^2))/(xsin(x^2))# Recalling trigonometric identities: #cota=cosa/sina# Let's split the result a bit. #(dy)/(dx)=(2x^cancel(2)color(green)(cos(x^2)))/(cancel(x)color(green)(sin(x^2)))-cancel((sin(x^2)))/(xcancel(sin(x^2)))# #(dy)/(dx)=2xcot(x^2)-1/x#

Adam Adkins · Answer

undefined To differentiate $ f(x) = \ln\left(\frac{\sin(x^2)}{x}\right) $, we can use the chain rule along with the quotient rule. The chain rule states that if we have a function within another function, we differentiate the outer function first and then multiply it by the derivative of the inner function.

First, we differentiate the outer function:

$$ \frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx} $$

Here, $ u = \frac{\sin(x^2)}{x} $.

Next, we find the derivative of $ u $ using the quotient rule:

$$ \frac{d}{dx}\left(\frac{\sin(x^2)}{x}\right) = \frac{x \cdot \cos(x^2) - \sin(x^2) \cdot 1}{x^2} $$

Now, we can substitute these derivatives back into the chain rule formula:

$$ f'(x) = \frac{1}{\frac{\sin(x^2)}{x}} \cdot \frac{x \cdot \cos(x^2) - \sin(x^2)}{x^2} $$

Simplifying further:

$$ f'(x) = \frac{x \cdot \cos(x^2) - \sin(x^2)}{x \cdot \sin(x^2)} $$

This is the derivative of $ f(x) = \ln\left(\frac{\sin(x^2)}{x}\right) $.