# Use differentiation Rules to find the following derivative: `((e^(2x)-x)/(4cosx))' = ?`

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You need to differentiate the given rational function using the following rules, such that:

`((e^(2x) - x)/(4 cos x))' = ((e^(2x) - x)'(4 cos x) - (e^(2x) - x)(4 cos x)')/(16 cos^2 x)` (quotient rule)

`((e^(2x) - x)/(4 cos x))' = ((e^(2x)*(2x)' - 1)(4 cos x) - (e^(2x) - x)(4 (-sin x)))/(16 cos^2 x)` (chain rule used to differentiate `e^(2x))`

`((e^(2x) - x)/(4 cos x))' = ((2e^(2x) - 1)(4cos x) + (4sin x)(e^(2x) - x))/(16 cos^2 x)`

You need to factor out 4 such that:

`((e^(2x) - x)/(4 cos x))' = 4((2e^(2x) - 1)(cos x) + (sin x)(e^(2x) - x))/(16 cos^2 x)`

Reducing duplicate factors yields:

`((e^(2x) - x)/(4 cos x))' = ((2e^(2x) - 1)(cos x) + (sin x)(e^(2x) - x))/(4 cos^2 x)`

**Hence, evaluating the derivative of the given rational function yields `((e^(2x) - x)/(4 cos x))' = ((2e^(2x) - 1)(cos x) + (sin x)(e^(2x) - x))/(4 cos^2 x).` **

Let the expression be:

`y =(e^(2x)-x)/(4cosx)`

To take the derivative of y, use the quotient rule (u/v)' = (vu'-uv')/v^2

`y' =((e^(2x)-x)/(4cosx))'`

So,

`u= e^(2x) - x ` and `v = 4cos x`

Then, take the derivative of u and v .

As per exponential formula of derivative which is `(e^u)' =e^u*u'` and the basic formula `(x)' = 1` , then,

`u ' = e^(2x) * 2 - 1 = 2e^(2x) - 1`

And to determine v', apply the trigonometric formula `(cos u)' = -sin u * u'` .

`v ' = -4sinx * x' = -4sinx *1 = -4sinx`

Next, substitute u, v, u' and v' to the quotient formula of derivatives.

`y ' = ((4cosx) *(2e^(2x)-1)-(e^(2x)-x)*(-4sinx))/(4cosx)^2`

`y' = (8cosxe^(2x) - 4cosx -(-4e^(2x)sinx+4xsinx))/(16cos^2x)`

`y'=(8cosxe^(2x) - 4cosx+4e^(2x)sinx - 4xsinx)/(16cos^2x)`

Factor out the GCF at the numerator.

`y'=(4(2cosx e^(2x) - cosx + e^(2x)sinx-xsinx))/(16cos^2x)`

Cancel common factor between numerator and denominator to simplify.

`y' =(2cosx e^(2x) - cosx + e^(2x)sinx-xsinx)/(4cos^2x)`

**Hence, `((e^(2x) - x)/(4cosx))'= (2cosxe^(2x) - cosx + e^(2x)sinx-xsinx)/(4cos^2x)` .**