How do you differentiate f(x)=(xtanx)/(x^2-x+3) using the quotient rule?

Mar 29, 2018

$- x \cdot {\left({x}^{2} - x + 3\right)}^{-} 2 \cdot \left(2 x - 1\right) \cdot \tan x + {\left({x}^{2} - x + 3\right)}^{-} 1 \cdot \tan x + x \cdot {\left({x}^{2} - x + 3\right)}^{-} 1 \cdot {\sec}^{2} x$

There are 2 Answers first with the product rule the second with the quotient rule

Explanation:

you can write the function like this: $f \left(x\right) = x \cdot {\left({x}^{2} - x + 3\right)}^{-} 1 \cdot \tan x$

now consider:
$g \left(x\right) = x$

$h \left(x\right) = \tan x$
$v \left(x\right) = {\left({x}^{2} - x + 3\right)}^{-} 1$

$\frac{d}{\mathrm{dx}} \left(g \left(x\right) \cdot h \left(x\right) \cdot v \left(x\right)\right) = g ' \left(h\right) \cdot h \left(x\right) \cdot v \left(x\right) + g \left(x\right) \cdot h \left(x\right) \cdot v ' \left(x\right) + g \left(x\right) \cdot h ' \left(x\right) \cdot v \left(x\right)$

$g ' \left(x\right) = 1$

$h ' \left(x\right) = {\sec}^{2} x$

$v ' \left(x\right) = - {\left({x}^{2} - x + 3\right)}^{-} 2 \cdot \left(2 x - 1\right)$

and by substituting in the relation above you get:
$\frac{d}{\mathrm{dx}} x \cdot {\left({x}^{2} - x + 3\right)}^{-} 1 \cdot \tan x = - x \cdot {\left({x}^{2} - x + 3\right)}^{-} 2 \cdot \left(2 x - 1\right) \cdot \tan x + {\left({x}^{2} - x + 3\right)}^{-} 1 \cdot \tan x + x \cdot {\left({x}^{2} - x + 3\right)}^{-} 1 \cdot {\sec}^{2} x$

y'=(g(x)f'(x)−f(x)g'(x))/(g(x))^2
where $f \left(x\right) = x \tan x$
$g \left(x\right) = {x}^{2} - x + 3$
$f ' \left(x\right) = x {\sec}^{2} x + \tan x$
$g ' \left(x\right) = 2 x - 1$
$y ' = \frac{\left({x}^{2} - x + 3\right) \left(x {\sec}^{2} x + \tan x\right) - \left(x \tan x\right) \left(2 x - 1\right)}{{x}^{2} - x + 3} ^ 2$