How do you differentiate #f(x) = (cos2x)/(e^(2x)-x)# using the quotient rule?

Question

Christopher Agresta · Accepted Answer

#(-2sin(2x)*(e^(2x)-x)-(2e^(2x)-1)*cos(2x))/(e^(2x)-x)^2#

The quotient rule states that, given #(f/g)'#, (mind that this is in Newton's notation), the derivative is #(f'g-fg')/g^2#

Here, #f=cos2x#, and #g=e^(2x)-x#.

Let us approach the issue gradually.

First, #f'#.

The derivative of #cos2x# cannot be calculated directly. We must use the chain rule, which states that #(df(u))/dx=(df)/(du)*(du)/dx#.

Here, #f=cosu#, and #u=2x#

The derivative of #cosu=-sinu#, and the derivative of #2x=2#.

Multiplying: #-2sinu#

#f'=-2sin2x#

Next, #g'#.

Again, the derivative of #e^2x-x# cannot be done directly. The difference rule states that #(f-g)'=f'-g'#.

So first, #d/dxe^(2x)#. Use the chain rule again, here #f=e^u# and #g=2x#.

The derivative of #e^u=e^u#, and of #2x=2#

So this becomes #2e^u#.

#g'=2e^(2x)-1#. The #1# is the derivative of #x#.

We can now just enter.

#(f'g-fg')/g^2#

#((-2sin2x)(e^(2x)-x)-(cos2x)(2e^(2x)-1))/(e^(2x)-x)^2#

#(-2sin(2x)*(e^(2x)-x)-(2e^(2x)-1)*cos(2x))/(e^(2x)-x)^2#

Cameron Alexander · Answer

undefined To differentiate $ f(x) = \frac{\cos^2(2x)}{e^{2x} - x} $ using the quotient rule, you would apply the formula:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Where $ u(x) = \cos^2(2x) $ and $ v(x) = e^{2x} - x $. Then, differentiate each function separately:

$$ u'(x) = -4\sin(2x)\cos(2x) $$
$$ v'(x) = 2e^{2x} - 1 $$

Finally, plug these into the quotient rule formula to get the derivative of $ f(x) $:

$$ f'(x) = \frac{-4\sin(2x)\cos(2x)(e^{2x} - x) - \cos^2(2x)(2e^{2x} - 1)}{(e^{2x} - x)^2} $$