How do you use the quotient rule to differentiate #y=cos(x)/ln(x)#?

1 Answer
Aug 4, 2014

The quotient rule states:

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

Let #f(x) = cosx#, and let #g(x) = ln x#.

We know that the derivative of #cosx# is #-sinx#, and that the derivative of #ln x# is #1/x#. Therefore, #f'(x) = -sinx#, and #g'(x) = 1/x#.

Now we may simply plug into the quotient rule formula:

#dy/dx = (-sinxlnx - cosx/x)/((ln x)^2)#

And there is our answer. If we would like, we can split this fraction up to make it a bit prettier:

#dy/dx = -(sinxlnx)/((ln x)^2) - (cosx/x)/((ln x)^2)#

This simplifies to:

#dy/dx = -(sinx)/(ln x) - cosx/(x(ln x)^2)#