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Calculus

posted by on .

Differentiate the following function:
y= [x+1]/√x

I tried
y=[x+1]/x^(1/2)
y' =1/2√x
y'=2√x

Thanks in advance.

  • Calculus - ,

    OH MY!

    my first line after the quotient rule is
    dy/dx = (√x - (1/2)x(-1/2)(x+1))/x

    which reduces to (x+1)/(2x^(3/2))

  • Calculus - ,

    What did you do to get the first line?

  • Calculus - ,

    As I said, I used the quotient rule.

    Judging by the type of question you are differentiating, you must know that.

    alternate way,
    change your question to
    y = (x+1)(x^(-1/2)) and use the product rule.
    same result of course.

  • Calculus - ,

    Oh, I haven't learnt the quotient rule, only the power rule, thanks.

  • Calculus - ,

    ok then try this

    y = (x+1)/√x
    = x/√x + 1/√x
    = x^(1/2) + x^(-1/2)

    so dy/dx = (1/2)x(-1/2) - (1/2)x^(-3/2)
    = (1/2)x^(-3/2)[x - 1}
    = (x-1)/(2x^(3/2))

    Just noticed that in my intial reply I had x+1 instead of x-1 in the numerator.
    This last result is the correct one.

  • Calculus - ,

    The back of the textbook just left it as y'= (1/2)x(-1/2) - (1/2)x^(-3/2)

    Thanks! :)

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