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

1 Answer
Guilherme N.
Dec 30, 2015

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#

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