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Continuity And Differentiability

Question
CBSEENMA12035636

Find space dy over dx space if space straight y equals sin to the power of negative 1 end exponent open square brackets fraction numerator 6 straight x minus 4 square root of 1 minus 4 straight x squared end root over denominator 5 end fraction close square brackets

Solution
Given space that
straight y space equals space sin to the power of negative 1 end exponent space open square brackets fraction numerator 6 straight x minus 4 square root of 1 space minus space 4 straight x squared end root over denominator 5 end fraction close square brackets

if space straight y space equals space sin to the power of negative 1 end exponent straight x comma space then space dy over dx space equals fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction

straight y space equals space sin to the power of negative 1 end exponent open square brackets fraction numerator 6 straight x minus 4 square root of 1 space minus space 4 straight x squared end root over denominator 5 end fraction close square brackets

rightwards double arrow space space straight y space equals space sin to the power of negative 1 end exponent open square brackets fraction numerator 6 straight x over denominator 5 end fraction minus fraction numerator 4 square root of 1 space minus space 4 straight x squared end root over denominator 5 end fraction close square brackets

rightwards double arrow space straight y space equals space sin to the power of negative 1 end exponent open square brackets fraction numerator 2 straight x.3 over denominator 3 end fraction space minus space fraction numerator 4 square root of 1 space minus space left parenthesis 2 straight x right parenthesis squared end root over denominator 5 end fraction close square brackets

rightwards double arrow space straight y space equals space sin to the power of negative 1 end exponent open square brackets 2 straight x.3 over 5 minus 4 over 5 square root of 1 space minus space left parenthesis 2 straight x right parenthesis squared end root close square brackets

rightwards double arrow space straight y space equals space sin to the power of negative 1 end exponent open square brackets 2 straight x space square root of 1 minus open parentheses 4 over 5 close parentheses squared end root minus space 4 over 5 square root of 1 space minus left parenthesis 2 straight x right parenthesis squared end root close square brackets

we space know space that comma

sin to the power of negative 1 end exponent straight p space minus space sin to the power of negative 1 end exponent straight q space equals space sin to the power of negative 1 end exponent space left parenthesis straight p square root of 1 space minus space straight q squared end root minus straight q-th root of 1 space minus space straight p squared end root right parenthesis

Here comma space straight p space equals space 2 straight x space space and space straight q space equals 4 over 5

Differentiating space the space above space functions space with space respect space straight x comma space space we space have comma space

dy over dx space equals space fraction numerator 1 over denominator square root of 1 minus left parenthesis 2 straight x right parenthesis squared end root end fraction space straight x space 2 minus 0

rightwards double arrow space space dy over dx space equals space fraction numerator 2 over denominator square root of 1 minus 4 straight x squared end root end fraction
space

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