Question

what is the derivative of the function f(x) = x^2+1 /2x

Answer 1

To find the derivative of the function \( f(x) = \frac{x^2 + 1}{2x} \), we can use the quotient rule, which states that if \( f(x) = \frac{g(x)}{h(x)} \), then the derivative \( f'(x) \) is given by:

\[
f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}
\]

In this case:
- \( g(x) = x^2 + 1 \)
- \( h(x) = 2x \)

First, we need to compute the derivatives \( g'(x) \) and \( h'(x) \):

1. \( g'(x) = 2x \)
2. \( h'(x) = 2 \)

Now we apply the quotient rule:

\[
f'(x) = \frac{(2x)(2x) - (x^2 + 1)(2)}{(2x)^2}
\]

Simplifying the numerator:

1. Calculate \( (2x)(2x) = 4x^2 \).
2. Calculate \( (x^2 + 1)(2) = 2x^2 + 2 \).

Now substituting back into the formula:

\[
f'(x) = \frac{4x^2 - (2x^2 + 2)}{(2x)^2}
\]

Simplifying the numerator:

\[
4x^2 - 2x^2 - 2 = 2x^2 - 2
\]

Thus, we have:

\[
f'(x) = \frac{2x^2 - 2}{(2x)^2} = \frac{2(x^2 - 1)}{4x^2} = \frac{x^2 - 1}{2x^2}
\]

So the final derivative is:

\[
f'(x) = \frac{x^2 - 1}{2x^2}
\]