# Question b4941

Nov 10, 2017

The quotient rule is:

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

In this case, $u = {x}^{2} + {x}^{2} {\tan}^{2} \left(x\right)$ and $v = {\sec}^{2} \left(x\right)$, then

$u ' = 2 x + 2 x {\tan}^{2} \left(x\right) + 2 {x}^{2} \tan \left(x\right) {\sec}^{2} \left(x\right)$ and $v ' = 2 \tan \left(x\right) {\sec}^{2} \left(x\right)$

Substituting these values into the quotient rule:

$\frac{d \left(\frac{{x}^{2} + {x}^{2} {\tan}^{2} \left(x\right)}{\sec} ^ 2 \left(x\right)\right)}{\mathrm{dx}} = \frac{\left(2 x + 2 x {\tan}^{2} \left(x\right) + 2 {x}^{2} \tan \left(x\right) {\sec}^{2} \left(x\right)\right) {\sec}^{2} \left(x\right) - \left({x}^{2} + {x}^{2} {\tan}^{2} \left(x\right)\right) \left(2 \tan \left(x\right) {\sec}^{2} \left(x\right)\right)}{\sec} ^ 4 \left(x\right)$

A common factor of ${\sec}^{2} \frac{x}{\sec} ^ 2 \left(x\right)$ becomes 1:

$\frac{d \left(\frac{{x}^{2} + {x}^{2} {\tan}^{2} \left(x\right)}{\sec} ^ 2 \left(x\right)\right)}{\mathrm{dx}} = \frac{\left(2 x + 2 x {\tan}^{2} \left(x\right) + 2 {x}^{2} \tan \left(x\right) {\sec}^{2} \left(x\right)\right) - \left({x}^{2} + {x}^{2} {\tan}^{2} \left(x\right)\right) \left(2 \tan \left(x\right)\right)}{\sec} ^ 2 \left(x\right)$

The term $- \left({x}^{2} + {x}^{2} {\tan}^{2} \left(x\right)\right) \left(2 \tan \left(x\right)\right)$ can be written as -2x^2tan(x)sec^2(x))#:

$\frac{d \left(\frac{{x}^{2} + {x}^{2} {\tan}^{2} \left(x\right)}{\sec} ^ 2 \left(x\right)\right)}{\mathrm{dx}} = \frac{\left(2 x + 2 x {\tan}^{2} \left(x\right) + 2 {x}^{2} \tan \left(x\right) {\sec}^{2} \left(x\right)\right) - 2 {x}^{2} \tan \left(x\right) {\sec}^{2} \left(x\right)}{\sec} ^ 2 \left(x\right)$

The last two terms in the numerator sum to 0:

$\frac{d \left(\frac{{x}^{2} + {x}^{2} {\tan}^{2} \left(x\right)}{\sec} ^ 2 \left(x\right)\right)}{\mathrm{dx}} = \frac{2 x + 2 x {\tan}^{2} \left(x\right)}{\sec} ^ 2 \left(x\right)$

Use the identity $1 + {\tan}^{2} \left(x\right) = {\sec}^{2} \left(x\right)$:

$\frac{d \left(\frac{{x}^{2} + {x}^{2} {\tan}^{2} \left(x\right)}{\sec} ^ 2 \left(x\right)\right)}{\mathrm{dx}} = 2 x {\sec}^{2} \frac{x}{\sec} ^ 2 \left(x\right)$

${\sec}^{2} \frac{x}{\sec} ^ 2 \left(x\right)$ becomes 1:

$\frac{d \left(\frac{{x}^{2} + {x}^{2} {\tan}^{2} \left(x\right)}{\sec} ^ 2 \left(x\right)\right)}{\mathrm{dx}} = 2 x$

It would have been better to avoid using the quotient rule by making the substitution, $1 + {\tan}^{2} \left(x\right) = {\sec}^{2} \left(x\right)$, at the start:

$\frac{{x}^{2} + {x}^{2} {\tan}^{2} \left(x\right)}{\sec} ^ 2 \left(x\right) = {x}^{2} {\sec}^{2} \frac{x}{\sec} ^ 2 \left(x\right) = {x}^{2}$

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