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Continuity And Differentiability

Question
CBSEENMA12035151
Wired Faculty App

Differerntiate space cos to the power of negative 1 end exponent open parentheses fraction numerator 3 cos space straight x minus 4 sin space straight x over denominator 5 end fraction close parentheses space straight w. straight r. straight t. straight x.

Solution
Let space space space space space straight y equals cos to the power of negative 1 end exponent open parentheses fraction numerator 3 cos space straight x minus 4 sin space straight x over denominator 5 end fraction close parentheses therefore space space space space space space space straight y equals cos to the power of negative 1 end exponent open parentheses 3 over 5 cos space straight x minus 4 over 5 sin space straight x close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis Put space 3 over 5 equals straight r space cosθ space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis and space 4 over 5 equals straight r space sinθ space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis squaring space and space adding space left parenthesis 2 right parenthesis space and space left parenthesis 3 right parenthesis comma space we space can space get comma space space space space 9 over 25 plus 16 over 25 equals straight r squared left parenthesis cos squared straight theta plus sin squared straight theta right parenthesis space rightwards double arrow space 1 equals straight r squared space rightwards double arrow straight r equals 1 Dividing space left parenthesis 3 right parenthesis space by space left parenthesis 2 right parenthesis comma space we space get comma space space space space space space space space space space space space space space space tan space straight theta equals 4 over 3 space rightwards double arrow space straight theta equals tan to the power of negative 1 end exponent space open parentheses 4 over 3 close parentheses therefore space from left parenthesis 1 right parenthesis comma space space straight y equals cos to the power of negative 1 end exponent left parenthesis straight r space cosθ space sin space straight x equals straight r space sinθ space sin space straight x right parenthesis space space space space space space space space space space space space space space space space space space space space space space space equals cos to the power of negative 1 end exponent left parenthesis cos space straight x space cosθ minus sin space xsinθ right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space straight r equals 1 right square bracket space space space space space space space space space space space space space space space space space space space space space space space equals cos to the power of negative 1 end exponent left square bracket cos left parenthesis straight x plus straight theta right parenthesis right square bracket equals straight x plus straight theta therefore space space space space space space space space space space space space space space space space space straight y equals straight x plus tan to the power of negative 1 end exponent 4 over 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space straight theta equals tan to the power of negative 1 end exponent 4 over 3 close square brackets therefore space space space space space space space space space space space space dy over dx equals 1 plus 0 space rightwards double arrow dy over dx equals 1

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