How do you differentiate f(x)=tan(x)/ (cos^2(x)cot (x)) using the quotient rule?

Dec 3, 2017

$\frac{2 \tan \left(x\right) + 2 {\sin}^{2} \left(x\right) \tan \left(x\right)}{\cos} ^ 4 \left(x\right)$

Explanation:

To avoid having to do first the quotient rule and then the product rule on the bottom, I will take advantage of the fact that $\cot \left(x\right) = \frac{1}{\tan} \left(x\right)$ and rewrite the function like this:
$\tan \frac{x}{{\cos}^{2} \left(x\right) \cot \left(x\right)} = \tan \frac{x}{{\cos}^{2} \frac{x}{\tan} \left(x\right)} = \tan \left(x\right) \cdot \tan \frac{x}{\cos} ^ 2 \left(x\right) = {\tan}^{2} \frac{x}{\cos} ^ 2 \left(x\right)$

The quotient rule states that:
$\frac{d}{\mathrm{dx}} \left(f \frac{x}{g} \left(x\right)\right) = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

In our case, $f \left(x\right) = {\tan}^{2} \left(x\right)$ and $g \left(x\right) = {\cos}^{2} \left(x\right)$. The derivatives are:
$f ' \left(x\right) = 2 \tan \left(x\right) {\sec}^{2} \left(x\right)$ (by the chain rule)

$g ' \left(x\right) = - 2 \cos \left(x\right) \sin \left(x\right)$ (again, by the chain rule)

Our derivative then becomes:
$\frac{2 \tan \left(x\right) {\sec}^{2} \left(x\right) {\cos}^{2} \left(x\right) - \left(- 2 \sin \left(x\right) \cos \left(x\right) {\tan}^{2} \left(x\right)\right)}{{\cos}^{4} \left(x\right)}$

To simplify this, I will use thee two identities:
$\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$
$\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right)$

$\frac{2 \tan \left(x\right) \frac{1}{\cancel{{\cos}^{2} \left(x\right)}} \cancel{{\cos}^{2} \left(x\right)} + 2 \sin \left(x\right) \cancel{\cos \left(x\right)} {\sin}^{2} \frac{x}{\cos} ^ \cancel{2} \left(x\right)}{\cos} ^ 4 \left(x\right)$

$\frac{2 \tan \left(x\right) + 2 {\sin}^{3} \frac{x}{\cos} \left(x\right)}{\cos} ^ 4 \left(x\right) = \frac{2 \tan \left(x\right) + 2 {\sin}^{2} \left(x\right) \tan \left(x\right)}{\cos} ^ 4 \left(x\right)$