find the derivative of f(x)=cot(3x)csc^2(3x)
f(x)=cot(3x)csc^2(3x) = cosx/sin^3x
Let sinx = u . Then du = cosx dx
The integral of f(x) becomes
Integral of du/u^3 = -(1/2) u^-2
= (-1/2)(1/sinx)^2
= (-1/2)csc^2x
first change it to
f(x) = cos 3x/(sin 3x)^3 like drwls had
now use the quotient rule to find f'(x)
f'(x) = [(sin 3x)^3(-3)(sin 3x) - 3(sin 3x)^2(cos 3x)(3)(cos 3x)]/(sin 3x)^6
= -3[(sin 3x)^2 + 3(cos 3x)^2]/(sin 3x)^4
I made at least one mistake. The first was using x instead of 3x.
To find the derivative of the function f(x) = cot(3x)csc^2(3x), we can use the product rule and the chain rule.
Let's first rewrite the function as f(x) = cot(3x)(csc(3x))^2.
The product rule states that if we have a function f(x) = g(x) * h(x), then the derivative of f(x) can be found using the formula:
f'(x) = g'(x) * h(x) + g(x) * h'(x).
Using the product rule, let's find the derivative of f(x):
Step 1: Find the derivative of g(x) = cot(3x).
The derivative of cot(x) is -csc^2(x), so the derivative of cot(3x) is -3csc^2(3x).
Step 2: Find the derivative of h(x) = (csc(3x))^2.
To find the derivative of (csc(3x))^2, we can use the chain rule. Let u = csc(3x), then the function can be rewritten as f(x) = u^2. Taking the derivative of f(x) with respect to x using the chain rule, we get:
h'(x) = (2u) * (du/dx).
To find du/dx, we differentiate u = csc(3x). The derivative of csc(x) is -csc(x)cot(x), so the derivative of csc(3x) is -3csc(3x)cot(3x). Thus,
du/dx = -3csc(3x)cot(3x).
Substituting this back into h'(x), we have h'(x) = (2u) * (-3csc(3x)cot(3x)) = -6csc(3x)cot(3x)u = -6csc(3x)cot(3x)(csc(3x))^2.
Step 3: Apply the product rule formula to find f'(x):
f'(x) = g'(x) * h(x) + g(x) * h'(x)
= (-3csc^2(3x)) * (csc(3x))^2 + cot(3x) * (-6csc(3x)cot(3x)(csc(3x))^2)
= -3csc^2(3x) * csc^2(3x) - 6csc(3x)cot^2(3x)(csc(3x))^2.
Simplifying further, we have:
f'(x) = -3(csc^2(3x))^2 - 6csc(3x)cot^2(3x)(csc^2(3x))
= -3csc^4(3x) - 6csc(3x)cot^2(3x)csc^2(3x)
= -3csc^4(3x) - 6cot^2(3x)csc^3(3x).
Therefore, the derivative of f(x) = cot(3x)csc^2(3x) is f'(x) = -3csc^4(3x) - 6cot^2(3x)csc^3(3x).