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If  f(x) = open parentheses fraction numerator a plus x over denominator b plus x end fraction close parentheses to the power of a plus b plus x end exponent     Prove that f' (0) = open parentheses a over b close parentheses to the power of a plus b space end exponent space open square brackets space L o g space a over b space plus fraction numerator space b squared space minus space a squared space space space over denominator a b end fraction space space space space space space close square brackets

Asked by ahuja8087 17th March 2017, 5:28 PM
Answered by Expert
Answer:

begin mathsize 16px style straight f left parenthesis straight x right parenthesis equals open parentheses fraction numerator straight a plus straight x over denominator straight b plus straight x end fraction close parentheses to the power of straight a plus straight b plus straight x end exponent
Take space log space on space both space sides comma space we space get
log space straight f left parenthesis straight x right parenthesis space equals space left parenthesis straight a plus straight b plus straight x right parenthesis space log open parentheses fraction numerator straight a plus straight x over denominator straight b plus straight x end fraction close parentheses
log space straight f left parenthesis straight x right parenthesis space equals space left parenthesis straight a plus straight b plus straight x right parenthesis space open square brackets log open parentheses straight a plus straight x close parentheses minus log left parenthesis straight b plus straight x right parenthesis close square brackets
rightwards double arrow fraction numerator 1 over denominator straight f left parenthesis straight x right parenthesis end fraction straight f apostrophe left parenthesis straight x right parenthesis equals left parenthesis straight a plus straight b plus straight x right parenthesis space straight d over dx space open square brackets log open parentheses straight a plus straight x close parentheses minus log left parenthesis straight b plus straight x right parenthesis close square brackets plus space open square brackets log open parentheses straight a plus straight x close parentheses minus log left parenthesis straight b plus straight x right parenthesis close square brackets straight d over dx left parenthesis straight a plus straight b plus straight x right parenthesis
Simplify space and space then space replace space straight x space by space 0 space througout. end style

Answered by Expert 17th March 2017, 5:45 PM
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