Asked by redgy
find f'(x) for f(x) = x/(sqrtx^2+1) please show all the steps. I'm realy confused on this one. I get as far as setting it up using the quotient rule then I go blank. thanks
Answers
Answered by
Reiny
I will assume you meanst
f(x) = x/√(x^2+1) or x/(x^2 + 1)^(1/2)
did you have
f'(x) = [(x^2 + 1)^(1/2) - x(1/2)(x^2 + 1)^(-1/2)(2x)]/(x^2 + 1) ?
take out a common factor of (x^2+1)^(-1/2) from the numerator
= (x^2 + 1)^(-1/2) [ x^2 + 1 - x^2] / (x^2 + 1)
= 1/(x^2 + 1)^(3/2) or (x^2 + 1)^(-3/2)
f(x) = x/√(x^2+1) or x/(x^2 + 1)^(1/2)
did you have
f'(x) = [(x^2 + 1)^(1/2) - x(1/2)(x^2 + 1)^(-1/2)(2x)]/(x^2 + 1) ?
take out a common factor of (x^2+1)^(-1/2) from the numerator
= (x^2 + 1)^(-1/2) [ x^2 + 1 - x^2] / (x^2 + 1)
= 1/(x^2 + 1)^(3/2) or (x^2 + 1)^(-3/2)
Answered by
redgy
THANK YOU!!!!!!
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