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.
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! :)
To differentiate the function y = (x+1)/√x, you can use the quotient rule.
The quotient rule states that if you have a function in the form f(x) = g(x)/h(x), where g(x) and h(x) are two differentiable functions, then the derivative f'(x) can be found using the formula:
f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))^2
In this case, g(x) = (x+1) and h(x) = √x.
Now let's find the derivatives of g(x) and h(x):
g'(x) = 1 (the derivative of x+1 is 1)
h'(x) = (1/2)x^(-1/2) (the derivative of √x is (1/2)x^(-1/2))
Now we can plug these values into the quotient rule formula:
f'(x) = [(1)(√x) - (x+1)((1/2)x^(-1/2))] / (√x)^2
= (√x - (1/2)(x+1)(x^(-1/2))) / x
= (√x - (1/2)(x+1)/√x) / x
= (√x - (1/2)(√x + 1/√x)) / x
= (√x - (1/2)(2√x)) / x
= (√x - √x) / x
= 0 / x
= 0
Therefore, the derivative of y = (x+1)/√x is y' = 0.