How do you differentiate #f(x)=ln(sinx)/cosx# using the quotient rule?

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Answer 1 · 2016-02-08T04:19:11

#f'(x)=(cosxcotx+sinxln(sinx))/cos^2x#

The quotient rule states that

#d/dx[g(x)/(h(x))]=(g'(x)h(x)-g(x)h'(x))/[h(x)]^2#

When this is applied to #f(x)#, the function at hand, we see that

#f'(x)=(cosxd/dx[ln(sinx)]-ln(sinx)d/dx[cosx])/cos^2x#

Find each derivative separately:

To differentiate the natural logarithm function, use the chain rule, which for a natural logarithm function states that

#d/dx[ln(k(x))]=1/(k(x))*k'(x)#

Here, #k(x)=sinx#, so

#d/dx[ln(sinx)]=1/sinx*d/dx[sinx]=1/sinx*cosx=cotx#

As for the other derivative,

#d/dx[cosx]=-sinx#

Plugging these both back in, we see that

#f'(x)=(cosx(cotx)-ln(sinx)(-sinx))/cos^2x#

This can be written as

#f'(x)=(cosxcotx+sinxln(sinx))/cos^2x#

As with many trigonometric functions, this can be rewritten in many ways, including

#f'(x)=secx(cotx+tanxln(sinx))#