Using the quotient rule find the first derivative of the function y=((3x+1)(cos2x))/e^2x

To find the derivative of the function y = ((3x+1)(cos2x))/e^(2x) using the quotient rule, we need to find the derivatives of the numerator and denominator separately.

First, let's find the derivative of the numerator using the product rule:

dy/dx = (d/dx)[(3x+1)(cos2x)]
= (3)(cos2x) + (3x+1)(d/dx)(cos2x)

Using the chain rule, the derivative of cos2x is -2sin2x:

dy/dx = 3cos2x + (3x+1)(-2sin2x)

Next, let's find the derivative of the denominator:

dy/dx = (d/dx)[e^(2x)]
= e^(2x)(d/dx)(2x)

Using the chain rule, the derivative of 2x is 2:

dy/dx = e^(2x)(2)

Now, we can apply the quotient rule:

dy/dx = (denominator * derivative of numerator - numerator * derivative of denominator) / denominator^2

dy/dx = (e^(2x)(2)[(3cos2x) + (3x+1)(-2sin2x)] - [(3x+1)(cos2x)](2e^(2x))) / (e^(2x))^2

Simplifying the expression:

dy/dx = (2e^(2x)[3cos2x - (3x+1)(2sin2x)] - 2(3x+1)(cos2x)e^(2x)) / e^(4x)

dy/dx = (2e^(2x)(3cos2x - (3x+1)(2sin2x)) - 2(3x+1)(cos2x)e^(2x)) / e^(4x)

Therefore, the first derivative of the given function y = ((3x+1)(cos2x))/e^(2x) is (2e^(2x)(3cos2x - (3x+1)(2sin2x)) - 2(3x+1)(cos2x)e^(2x)) / e^(4x).

To find the first derivative of the function y=((3x+1)(cos2x))/e^2x using the quotient rule, we need to differentiate the numerator and denominator separately and apply the quotient rule formula.

The quotient rule states that for a function f(x) = u(x)/v(x), its derivative is given by:

f'(x) = (u'(x)v(x) - u(x)v'(x)) / v(x)^2

Let's apply this rule to the given function step-by-step:

Step 1: Determine u(x) and v(x)
u(x) = (3x + 1)(cos2x)
v(x) = e^(2x)

Step 2: Find the derivatives of u(x) and v(x)
u'(x) = (3)(cos2x) + (3x + 1)(-2sin2x)
= 3cos2x - 6xsin2x - 2sin2x

v'(x) = 2e^(2x) (since the derivative of e^ax is ae^ax)

Step 3: Apply the quotient rule formula
f'(x) = (u'(x)v(x) - u(x)v'(x)) / v(x)^2
= ((3cos2x - 6xsin2x - 2sin2x)e^(2x) - (3x + 1)(cos2x)(2e^(2x))) / (e^(2x))^2

Simplifying the expression, we have:

f'(x) = (3cos2x - 6xsin2x - 2sin2x)e^(2x) - 2(3x + 1)(cos2x)e^(2x) / e^(4x)

Therefore, the first derivative of the function y=((3x+1)(cos2x))/e^2x is:

f'(x) = (3cos2x - 6xsin2x - 2sin2x)e^(2x) - 2(3x + 1)(cos2x)e^(2x) / e^(4x)