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.

6 answers

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))
What did you do to get the first line?
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.
Oh, I haven't learnt the quotient rule, only the power rule, thanks.
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.
The back of the textbook just left it as y'= (1/2)x(-1/2) - (1/2)x^(-3/2)

Thanks! :)
Similar Questions
  1. Differentiate each functiona) y = -3x^2 + 5x - 4 b) f(x) = 6/x - 3/(x^2) c) f(x) = (3x^2 - 4x)(x^3 + 1) thank you in advance :)
    1. answers icon 2 answers
  2. 1. differentiate cos(3/x)2. differentiate sin(4/x) 3. differentiate 3/{sin(3x+pi)} 4. differentiate pxsin(q/x)where p and q are
    1. answers icon 0 answers
    1. answers icon 0 answers
    1. answers icon 2 answers
more similar questions