Differentiate.
f(x) = 4xe^xcsc(x)
Just use the product rule
f = 4x e^x cscx
f' = (4)(e^x cscx) + (4x)(e^x)(cscx) + (4x e^x)(-cscx cotx)
= -4e^x cscx (x cotx - x - 1)
To differentiate the function f(x) = 4xe^xcsc(x), we can use the product rule and the chain rule.
Step 1: Apply the product rule
The product rule states that if f(x) = u(x)v(x), then its derivative f'(x) can be calculated as follows: f'(x) = u'(x)v(x) + u(x)v'(x).
Let's assign u(x) = 4x and v(x) = e^xcsc(x).
Step 2: Find the derivatives of u(x) and v(x)
To differentiate u(x) = 4x, we can use the power rule. The power rule states that if g(x) = x^n, then its derivative g'(x) = nx^(n-1).
So, u'(x) = 4.
To differentiate v(x) = e^xcsc(x), we need to use the chain rule. The chain rule states that if y = f(u(x)), then dy/dx = f'(u(x)) * u'(x).
Let's differentiate f(x) = ex first. The derivative of f(x) = ex is simply f'(x) = ex.
Now, let's differentiate g(x) = csc(x). The derivative of g(x) = csc(x) is g'(x) = -csc(x)cot(x). (This can be derived using trigonometric identities).
Using the chain rule, we have v'(x) = ex * (-csc(x)cot(x)) = -excsc(x)cot(x).
Step 3: Apply the product rule formula
Now that we have obtained u'(x) and v'(x), we can substitute them into the formula f'(x) = u'(x)v(x) + u(x)v'(x).
Thus, f'(x) = (4)(e^xcsc(x)) + (4x)(-excsc(x)cot(x)).
Simplifying, we have f'(x) = 4e^xcsc(x) - 4excsc(x)cot(x).
So, the derivative of f(x) = 4xe^xcsc(x) is f'(x) = 4e^xcsc(x) - 4excsc(x)cot(x).