Posted by **sh** on Tuesday, October 20, 2009 at 10:53pm.

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 -
**Reiny**, Tuesday, October 20, 2009 at 10:59pm
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 -
**sh**, Tuesday, October 20, 2009 at 11:03pm
What did you do to get the first line?

- Calculus -
**Reiny**, Tuesday, October 20, 2009 at 11:07pm
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 -
**sh**, Tuesday, October 20, 2009 at 11:08pm
Oh, I haven't learnt the quotient rule, only the power rule, thanks.

- Calculus -
**Reiny**, Tuesday, October 20, 2009 at 11:35pm
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 -
**sh**, Tuesday, October 20, 2009 at 11:44pm
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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