# Question #9c03c

Dec 15, 2017

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

#### Explanation:

$f \left(x\right) = \tan \left(2 x + 3\right)$

The chain rule states that:

$\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Setting $u = 2 x + 3$, we get the equation:

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{d}{\mathrm{du}} \tan \left(u\right) \cdot \frac{d}{\mathrm{dx}} 2 x + 3$

And since $\frac{d}{\mathrm{du}} \tan \left(u\right)$ = ${\sec}^{2} \left(u\right)$ and $\frac{d}{\mathrm{dx}} 2 x + 3$ = 2:

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = {\sec}^{2} \left(u\right) \cdot 2$, which is:

$= {\sec}^{2} \left(2 x + 3\right) \cdot 2$

Dec 15, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {\sec}^{2} \left(2 x + 3\right)$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right)$

$y = \tan \left(2 x + 3\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \left(2 x + 3\right) \times \frac{d}{\mathrm{dx}} \left(2 x + 3\right)$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = 2 {\sec}^{2} \left(2 x + 3\right)$