# Calculus

posted by .

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

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

• 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,
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))

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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