# How do you differentiate f(x)= (2 x^2 + 7 x - 2)/ (x cos x ) using the quotient rule?

Jun 20, 2016

See below.

#### Explanation:

This will also require a use of the product rule for the function in the denominator.

The quotient rule states that for some function $h \left(x\right)$ that can be expressed as $f \frac{x}{g} \left(x\right)$, $h ' \left(x\right)$ is given by $\frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

Using that idea we are going to need to find $f ' \left(x\right)$ and $g ' \left(x\right)$:
$f ' \left(x\right)$ is just $4 x + 7$ and $g ' \left(x\right)$ is $\cos x - x \sin x$

Substituting into the rule, the derivative is equal to:
$\frac{\left(4 x + 7\right) \left(x \cos x\right) - \left(2 {x}^{2} + 7 x - 2\right) \left(\cos x - x \sin x\right)}{x \cos x} ^ 2$

You can then simplify as required.

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{\left(2 {x}^{2} + 2\right) \cdot \cos x + \left(2 {x}^{2} + 7 x - 2\right) x \cdot \sin x}{x \cdot \cos x} ^ 2$

#### Explanation:

The given equation is

$f \left(x\right) = \frac{2 {x}^{2} + 7 x - 2}{x \cdot \cos x}$

We use the formula for derivative of rational expression

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} \left(u\right) - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{v} ^ 2$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(\frac{2 {x}^{2} + 7 x - 2}{x \cdot \cos x}\right) = \frac{\left(x \cdot \cos x\right) \cdot \frac{d}{\mathrm{dx}} \left(2 {x}^{2} + 7 x - 2\right) - \left(2 {x}^{2} + 7 x - 2\right) \cdot \frac{d}{\mathrm{dx}} \left(x \cdot \cos x\right)}{x \cdot \cos x} ^ 2$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{\left(x \cdot \cos x\right) \cdot \left(4 x + 7\right) - \left(2 {x}^{2} + 7 x - 2\right) \left(- x \cdot \sin x + \cos x\right)}{x \cdot \cos x} ^ 2$

We simplify at this point

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{\left(4 {x}^{2} \cdot \cos x + 7 x \cdot \cos x\right) - \left(- 2 {x}^{3} \cdot \sin x - 7 {x}^{2} \cdot \sin x + 2 x \cdot \sin x + 2 {x}^{2} \cdot \cos x + 7 x \cdot \cos x - 2 \cdot \cos x\right)}{x \cdot \cos x} ^ 2$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{4 {x}^{2} \cdot \cos x + 7 x \cdot \cos x + 2 {x}^{3} \cdot \sin x + 7 {x}^{2} \cdot \sin x - 2 x \cdot \sin x - 2 {x}^{2} \cdot \cos x - 7 x \cdot \cos x + 2 \cdot \cos x}{x \cdot \cos x} ^ 2$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{\left(2 {x}^{2} + 2\right) \cdot \cos x + \left(2 {x}^{2} + 7 x - 2\right) x \cdot \sin x}{x \cdot \cos x} ^ 2$

God bless....I hope the explanation is useful.