How do you differentiate given sec^2(x) ?

Dec 9, 2016

$\frac{d {\sec}^{2} \left(x\right)}{\mathrm{dx}} = \frac{2 \sin \left(x\right)}{{\cos}^{3} \left(x\right)}$

Explanation:

Let $g \left(x\right) = \cos \left(x\right)$
and $f \left(x\right) = {x}^{- 2}$
so that
$\textcolor{w h i t e}{\text{XXX}} {\sec}^{2} \left(x\right) = \frac{1}{{\cos}^{2} \left(x\right)} = f \left(g \left(x\right)\right)$

We know that
$\textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \frac{d \cos \left(x\right)}{\mathrm{dx}} = - \sin \left(x\right)$
and
$\textcolor{w h i t e}{\text{XX}} \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{d {x}^{- 2}}{\mathrm{dx}} = - 2 {x}^{- 3}$
$\textcolor{w h i t e}{\text{XX}} \rightarrow \frac{d f \left(g \left(x\right)\right)}{d g \left(x\right)} = - 2 {g}^{- 3} \left(x\right) = - 2 {\cos}^{- 3} \left(x\right) = - \frac{2}{\cos \left(x\right)}$

By the Chain Rule:
$\textcolor{w h i t e}{\text{XXX}} \frac{d f \left(g \left(x\right)\right)}{\mathrm{dx}} = \frac{d f \left(g \left(x\right)\right)}{\mathrm{dx}} \times \frac{d g \left(x\right)}{d g \left(x\right)} = \frac{d f \left(g \left(x\right)\right)}{d g \left(x\right)} \times \frac{d g \left(x\right)}{\mathrm{dx}}$

Therefore
$\textcolor{w h i t e}{\text{XX}} \frac{d {\sin}^{2} \left(x\right)}{\mathrm{dx}}$
$\textcolor{w h i t e}{\text{XXXX}} = \frac{d f \left(g \left(x\right)\right)}{\mathrm{dx}} = - \frac{2}{\cos} ^ 3 \left(x\right) \times \left(- \sin \left(x\right)\right) = \frac{2 \sin \left(x\right)}{{\cos}^{3} \left(x\right)}$

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Although this is a different derivation than Bacterium's
the results are equivalent.

Some steps here are more explicit and it does not rely on remembering $\frac{d \sec \left(x\right)}{\mathrm{dx}} = \sec \left(x\right) \cdot \tan \left(x\right)$