find the derivative
f(x)= e^-3x x sin(4x)
To find the derivative of f(x) = e^(-3x) * x * sin(4x), we will use the product rule and the chain rule.
The product rule states that if we have two functions u(x) and v(x), the derivative of their product can be found using the formula (u * v)' = u' * v + u * v'. In this case, we have two functions: u(x) = e^(-3x) * x and v(x) = sin(4x).
First, let's find the derivative of u(x):
u'(x) = (e^(-3x) * x)'.
To find the derivative of e^(-3x), we need to apply the chain rule. The chain rule states that if we have a composition of functions g(f(x)), the derivative can be found by multiplying the derivative of the outer function g'(f(x)) and the derivative of the inner function f'(x).
Let's define g(f(x)) = e^x. In this case, f(x) = -3x and g(f(x)) = e^(-3x). The derivative of g(f(x)) = e^x is simply e^x.
Now we can find the derivative of e^(-3x) by applying the chain rule:
(e^(-3x))' = e^(-3x) * (-3) = -3e^(-3x).
Returning to u'(x):
u'(x) = -3e^(-3x) * x.
Now let's find the derivative of v(x):
v'(x) = (sin(4x))'.
The derivative of sin(4x) can be found using the chain rule. Let's define g(f(x)) = sin(x) and f(x) = 4x. The derivative of g(f(x)) = sin(x) is simply cos(x).
Applying the chain rule:
(sin(4x))' = cos(4x) * 4 = 4cos(4x).
Now, using the product rule, we can find the derivative of f(x):
f'(x) = u' * v + u * v'
= (-3e^(-3x) * x) * sin(4x) + (e^(-3x) * x) * 4cos(4x).
Simplifying the expression further, we have:
f'(x) = -3xe^(-3x)sin(4x) + 4xe^(-3x)cos(4x).
So, the derivative of f(x) = e^(-3x) * x * sin(4x) is f'(x) = -3xe^(-3x)sin(4x) + 4xe^(-3x)cos(4x).