Differentiate x/sqrt(x+a)?

Question

a ist a constant, should i bring up the square root of x+a with it being ^-1/2?

show me how you would do this

Answer 1

This question is used by the "quotient rule" in elementary calculus.

The formula is that the derivative of a fraction say f/g is (gf'-fg')/g²

So that will be the denominator sqrt(x+a) times the derivative of the numerator x which is 1 ... minus ... the numerator x times the derivative of the denominator sqrt(x+a) which is 1/2 times (x+a)^(-1/2)

that whole thing divided by sqrt(x+a) squared which is x+a

then simplify for your final answer

Answer 2

f(x) = x/sqrt(x + a)

It's perfectly valid to change sqrt(x + a) to (x + a)^(1/2).

f(x) = x / (x + a)^(1/2)

We could either use the quotient rule on this question, or, as you suggested, bring it up as (x + a)^(-1/2). I'm going to use the quotient rule.

f'(x) = [(1) (x + a)^(1/2) - x (1/[2(x + a)^(1/2)])] / (x + a)

The quotient rule is, verbally, "the derivative of the top times the bottom plus the derivative of the bottom times the top, over the bottom squared." Note that squaring the denominator is squaring a square root, meaning the answer is itself.

Simplifying further, we get

f'(x) = [(x + a)^(1/2) - x/[2(x + a)^(1/2)] / (x + a)

To get rid of the complex fraction, multiply top and bottom by
2(x + a).

f'(x) = [2(x + a)^(3/2) - x ] / (x + a)^(3/2)

We can stop here. On a final exam, they wouldn't normally ask you to simplify a derivative.

Answer 3

Yes you are right

x/sqrt(x+a) = x*(x+a)^(-1/2)

derivative is (x+a)^(-1/2) + x*(-1/2)*(x+a)^(-3/2) by the simple product rule, and as a fact product rule is computationally easier than the quotient rule all the time.

Whenever you need to take a derivative of a quotient, always convert it to a product by taking ^(-1) of the denominator. Its so much less simplification that way.

if you write this nicely its

1/sqrt(x+a) - x/ [2*(x+a)^(3/2)]