What is the derivative of #sqrt(x)/(x^3+1)# using the quotient rule?

Answer 1

We'll use the quotient rule, and, because the square root is used so often (for example: diagonal distance = hypotenuse of a right triangle), it's nice to memorize:

#d/(dx)sqrtx = 1/(2sqrtx)# (Go through it witl exponents to see that it's true.)

So:

#d/(dx)(sqrt(x)/(x^3+1)) = (1/(2sqrtx)(x^3+1) - (sqrtx) (3x^2))/((x^3+1)^2)#

Mulhtiply by #(2sqrtx)/(2sqrtx)# to get:

#d/(dx)(sqrt(x)/(x^3+1)) = ((x^3+1) - 2x (3x^2))/((2sqrtx)(x^3+1)^2) = (-5x^2+1)/((2sqrtx)(x^3+1)^2) #

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Answer 2

Samantha Adkison

To find the derivative of ( \frac{\sqrt{x}}{x^3 + 1} ) using the quotient rule:

Let ( u = \sqrt{x} ) and ( v = x^3 + 1 ).

Then, ( u' = \frac{1}{2\sqrt{x}} ) and ( v' = 3x^2 ).

Using the quotient rule, the derivative is given by:

[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} ]

Substitute the values:

[ \frac{d}{dx}\left(\frac{\sqrt{x}}{x^3 + 1}\right) = \frac{(x^3 + 1) \cdot \frac{1}{2\sqrt{x}} - \sqrt{x} \cdot 3x^2}{(x^3 + 1)^2} ]

[ = \frac{x^3 + 1 - 6x^2}{2\sqrt{x}(x^3 + 1)^2} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.