How do you find the derivative of #(2x^2 +x - 3)/x#?

Question

Emilia Aguilar · Accepted Answer

#2+3/x^2# #(2x^2+x-3)/x# #=(2x^2)/x+x/x-3/x# #=2x+1-3x^-1# differentiate using the #color(blue)"power rule"# #color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(ax^n)=nax^(n-1))color(white)(2/2)|)))# #rArrd/dx(2x+1-3x^-1)# #=2+3x^-2# #=2+3/x^2#

Evelyn Agard · Answer

undefined To find the derivative of $(2x^2 + x - 3)/x$, you can use the quotient rule. The quotient rule states that if you have a function $f(x) = \frac{g(x)}{h(x)}$, then its derivative is given by $$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}.$$

Using this rule, differentiate $(2x^2 + x - 3)/x$ with respect to $x$:

Let $g(x) = 2x^2 + x - 3$ and $h(x) = x$.

Now, differentiate $g(x)$ and $h(x)$ with respect to $x$:

$g'(x) = 4x + 1$
$h'(x) = 1$

Apply the quotient rule:

$$f'(x) = \frac{(4x + 1)(x) - (2x^2 + x - 3)(1)}{x^2}$$

$$= \frac{4x^2 + x - 2x^2 - x + 3}{x^2}$$

$$= \frac{2x^2 + 3}{x^2}$$

So, the derivative of $(2x^2 + x - 3)/x$ is $\frac{2x^2 + 3}{x^2}$.