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Continuity And Differentiability

Question
CBSEENMA12034854
Wired Faculty App

If space straight x to the power of straight p. straight y to the power of straight q equals left parenthesis straight x plus straight y right parenthesis to the power of straight p plus straight q end exponent comma space show space that space dy over dx equals straight y over straight x

Solution
space space space space space straight x to the power of straight p. straight y to the power of straight q equals left parenthesis straight x plus straight y right parenthesis to the power of straight p plus straight q end exponent rightwards double arrow space log left parenthesis straight x to the power of straight p. straight y to the power of straight q right parenthesis equals log left parenthesis straight x plus straight y right parenthesis to the power of straight p plus straight q end exponent rightwards double arrow space log space straight x to the power of straight p plus log space straight y to the power of straight q equals log left parenthesis straight x plus straight y right parenthesis to the power of straight p plus straight q end exponent rightwards double arrow straight p space log space straight x plus straight q space log space straight y equals left parenthesis straight p plus straight q right parenthesis log left parenthesis straight x plus straight y right parenthesis Differentiating space straight w. straight r. straight t. straight x comma space we space get comma space space space space space straight p over straight x plus straight q over straight y dy over dx equals fraction numerator straight p plus straight q over denominator straight x plus straight y end fraction open square brackets 1 plus dy over dx close square brackets therefore space straight p over straight x plus straight q over straight y dy over dx equals fraction numerator straight p plus straight q over denominator straight x plus straight y end fraction plus fraction numerator straight p plus straight q over denominator straight x plus straight y end fraction dy over dx space space space rightwards double arrow space space open parentheses straight q over straight y minus fraction numerator straight p plus straight q over denominator straight x plus straight y end fraction close parentheses dy over dx equals fraction numerator straight p plus straight q over denominator straight x plus straight y end fraction minus straight p over straight x therefore space open square brackets fraction numerator qz plus qy minus py minus qy over denominator straight y left parenthesis straight x plus straight y right parenthesis end fraction close square brackets dy over dx equals fraction numerator px plus qx minus px minus py over denominator straight x left parenthesis straight x plus straight y right parenthesis end fraction therefore space open square brackets fraction numerator qx minus py over denominator straight y end fraction close square brackets dy over dx equals fraction numerator qx minus py over denominator straight x end fraction space rightwards double arrow space space 1 over straight y dy over dx equals 1 over straight x therefore space dy over dx equals straight y over straight x.

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