Find dx/dt x = square root of 1+cot(3t)

Problem

d()/d(t)√(,1+cot(3*t))

Solution

1. Identify the outer function and the inner function to apply the Chain Rule. The expression is in the form u(1/2) where u=1+cot(3*t)

2. Differentiate the outer square root function using the Power Rule.

d(u(1/2))/d(u)=1/2*u(−1/2)

3. Apply the Chain Rule by multiplying by the derivative of the inner expression 1+cot(3*t)

d(x)/d(t)=1/(2√(,1+cot(3*t)))⋅d(1+cot(3*t))/d(t)

4. Differentiate the inner trigonometric function. The derivative of cot(u) is −csc2(u) and by the Chain Rule, we must also multiply by the derivative of 3*t

d(1+cot(3*t))/d(t)=−csc2(3*t)⋅3

5. Combine the results into a single expression.

d(x)/d(t)=(−3*csc2(3*t))/(2√(,1+cot(3*t)))

Final Answer

d(x)/d(t)=(−3*csc2(3*t))/(2√(,1+cot(3*t)))

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