How do you find the derivative of #(2sqrtx - 1)/(2sqrtx)#?

Question

Elijah Abram · Accepted Answer

#d/(dx) (frac (2sqrt(x)-1) (2sqrt(x)) )= 1/(4sqrt(x^3)#

We could use the quotient rule, stating that:

#d/(dx) ((f(x))/(g(x)) )= (f'(x)g(x)-f(x)g'(x))/(g(x)^2)#

It is however easier to write the function as:

# frac (2sqrt(x)-1) (2sqrt(x)) = 1-1/(2sqrt(x))=1-1/2x^(-1/2)#

so that:

#d/(dx) (frac (2sqrt(x)-1) (2sqrt(x)) )= d/(dx) ( 1-1/2x^(-1/2)) = 1/4x^(-3/2) =1/(4sqrt(x^3)#

Samantha Adkison · Answer

Break it into two fractions, the first becomes the derivative of a constant, and the second can be differentiated as a negative power.

Break into two fractions: #(d((2sqrt(x) - 1)/(2sqrt(x))))/dx = (d((2sqrt(x))/(2sqrt(x))))/dx - (d(1/(2sqrt(x))))/dx# Simplify: #(d((2sqrt(x) - 1)/(2sqrt(x))))/dx = (d(1))/dx - 1/2(d(x^(-1/2)))/dx# The derivative of a constant is zero and use the power rule on the second term: #(d((2sqrt(x) - 1)/(2sqrt(x))))/dx = 1/4x^(-3/2)#

Cameron Alexander · Answer

undefined To find the derivative of $\frac{2\sqrt{x} - 1}{2\sqrt{x}}$, you can use the quotient rule. The quotient rule states that if you have a function in the form $ \frac{f(x)}{g(x)}$, then its derivative is given by:

$$ \left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} $$

Let $f(x) = 2\sqrt{x} - 1$ and $g(x) = 2\sqrt{x}$. Then, $f'(x) = \frac{1}{\sqrt{x}}$ (using the power rule) and $g'(x) = \frac{1}{\sqrt{x}}$ (also using the power rule).

Now, substitute these values into the quotient rule formula:

$$ \left(\frac{2\sqrt{x} - 1}{2\sqrt{x}}\right)' = \frac{\left(\frac{1}{\sqrt{x}}\right)(2\sqrt{x}) - (2\sqrt{x} - 1)\left(\frac{1}{\sqrt{x}}\right)}{(2\sqrt{x})^2} $$

$$ = \frac{2 - (2\sqrt{x} - 1)}{4x} $$

$$ = \frac{2 - 2\sqrt{x} + 1}{4x} $$

$$ = \frac{3 - 2\sqrt{x}}{4x} $$