Question

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

Answer 1

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.

Explanation:

• Chain rule: `(dy)/(dx)=(dy)/(du)(du)/(dx)`
• Quotient rule: be `y=f(x)/g(x)`, then `y'=(f'g-fg')/g^2`

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`