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Math  /  Calculus

QuestionFind the derivative of the function. h(t)=11cot⁡−1(t)+11cot⁡−1(1t)h′(t)=□\begin{array}{l} h(t)=11 \cot ^{-1}(t)+11 \cot ^{-1}\left(\frac{1}{t}\right) \\ h^{\prime}(t)=\square \end{array}

Studdy Solution

STEP 1

1. The function to differentiate is h(t)=11cot⁡−1(t)+11cot⁡−1(1t) h(t) = 11 \cot^{-1}(t) + 11 \cot^{-1}\left(\frac{1}{t}\right) .
2. We need to find the derivative h′(t) h'(t) .

STEP 2

1. Differentiate 11cot⁡−1(t) 11 \cot^{-1}(t) .
2. Differentiate 11cot⁡−1(1t) 11 \cot^{-1}\left(\frac{1}{t}\right) .
3. Combine the derivatives to find h′(t) h'(t) .

STEP 3

Differentiate 11cot⁡−1(t) 11 \cot^{-1}(t) .
The derivative of cot⁡−1(t) \cot^{-1}(t) is −11+t2 -\frac{1}{1 + t^2} .
Thus, the derivative of 11cot⁡−1(t) 11 \cot^{-1}(t) is:
11(−11+t2)=−111+t2 11 \left(-\frac{1}{1 + t^2}\right) = -\frac{11}{1 + t^2}

STEP 4

Differentiate 11cot⁡−1(1t) 11 \cot^{-1}\left(\frac{1}{t}\right) .
Using the chain rule, the derivative of cot⁡−1(1t) \cot^{-1}\left(\frac{1}{t}\right) is:
−11+(1t)2⋅(−1t2) -\frac{1}{1 + \left(\frac{1}{t}\right)^2} \cdot \left(-\frac{1}{t^2}\right)
Simplify:
=11+1t2⋅1t2 = \frac{1}{1 + \frac{1}{t^2}} \cdot \frac{1}{t^2} =t2t2+1⋅1t2 = \frac{t^2}{t^2 + 1} \cdot \frac{1}{t^2} =1t2+1 = \frac{1}{t^2 + 1}
Thus, the derivative of 11cot⁡−1(1t) 11 \cot^{-1}\left(\frac{1}{t}\right) is:
11⋅1t2+1=11t2+1 11 \cdot \frac{1}{t^2 + 1} = \frac{11}{t^2 + 1}

STEP 5

Combine the derivatives to find h′(t) h'(t) .
h′(t)=−111+t2+11t2+1 h'(t) = -\frac{11}{1 + t^2} + \frac{11}{t^2 + 1}
Since the terms cancel each other out:
h′(t)=0 h'(t) = 0
The derivative of the function is:
0 \boxed{0}

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