Question

$Differentiate space cot to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses space straight w. straight r. straight t. straight x.$

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Solution

Let space space space space space straight y equals cot to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses Put space space space space space straight x equals tan space straight theta therefore space space space space space space space straight y equals cot to the power of negative 1 end exponent open parentheses fraction numerator 1 minus tan space straight theta over denominator 1 plus tan space straight theta end fraction close parentheses equals cot to the power of negative 1 end exponent open square brackets cot open parentheses straight pi over 4 plus straight theta close parentheses close square brackets equals straight pi over 4 plus straight theta equals straight pi over 4 plus tan to the power of negative 1 end exponent straight x therefore space dy over dx equals 0 plus fraction numerator 1 over denominator 1 plus straight x squared end fraction equals fraction numerator 1 over denominator 1 plus straight x squared end fraction

For Daily Free Study Material Join wiredfaculty