# Question 12c1e

Oct 23, 2017

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

#### Explanation:

The derivative of the function $\tan x$ can be found using the quotient rule, and remembering $\tan x = \frac{\sin x}{\cos x}$:

f(x) = (g(x))/(h(x)), f'(x) = (g'(x)h(x) - g(x)h'(x))/(h^2(x)

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

If we substitute $3 x + 2$ in place of $x$, then because $\frac{d}{\mathrm{dx}} \left(3 x + 2\right) = 3$, we multiply by 3 any portion where we took the derivative with respect to x, as indicated by the $\mathrm{dx}$ placed in the function above.

$\frac{d}{\mathrm{dx}} \tan \left(3 x + 2\right) = \left(\frac{3 {\cos}^{2} \left(3 x + 2\right) + 3 {\sin}^{2} \left(3 x + 2\right)}{{\cos}^{2} \left(3 x + 2\right)}\right) = 3 \frac{{\cos}^{2} \left(3 x + 2\right) + {\sin}^{2} \left(3 x + 2\right)}{{\cos}^{2} \left(3 x + 2\right)} = \frac{3 \cdot 1}{\cos} ^ 2 \left(3 x + 2\right) = 3 {\sec}^{2} \left(3 x + 2\right)$

We perform this last step by recalling our trig identities, namely that ${\sin}^{2} u + {\cos}^{2} u = 1$, where $u$ is any real, defined expression.

Oct 23, 2017

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

#### Explanation:

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

f’(x) = d/(dx) (tan (3x +2) )

u = (3x + 2); du= (3x+2) dx = 3

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

f’(x) = sec^2 u .du

Replacing value of u and du,
f’(x) = sec^2 (3x+2) * 3 = 3 sec^2(3x + 2)#