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))
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
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.
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.
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! :)
Thanks! :)