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

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).

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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) .

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