Question

Find the derivative of (3+tan X )( 4+3 Cos X)

Answer 1

answer: -9sinx + 3cosx + 4sec^2x

Answer 2

The derivative of $f(x) = (3 + \tan x)\cdot (4 + 3\cdot \cos x)$ is $f'(x) = \frac{4+3\cdot \cos x-9\cdot \sin x \cdot \cos^{2}x+3\cdot \sin^{2}x\cdot \cos x}{\cos^{2}x}$.

In this question we shall use the following derivative rules:

Derivative of a sum of functions

$\frac{d}{dt}(f(t) + g(t)) = \frac{df(t)}{dt} + \frac{dg(t)}{dt}$ (1)

Derivative of a product of functions

$\frac{d}{dt}(f(t)\cdot g(t)) = \frac{df(t)}{dt}\cdot g(t) + f(t)\cdot \frac{dg(t)}{dt}$ (2)

Derivative of a constant

$\frac{d}{dt}(k) = 0$, $\forall k \in \mathbb{R}$ (3)

Derivative of a function multiplied by a constant

$\frac{d}{dt}(c\cdot f(t)) = c\cdot \frac{df(t)}{dt}$, $\forall\, c\in \mathbb{R}$ (4)

Derivative of the cosine function

$\frac{d}{dt} \cos t = -\sin t$ (5)

Derivative of the tangent function

$\frac{d}{dt}\tan t = \sec^{2} t$ (6)

Chain rule

$\frac{d}{dt}f[u(t)] = \frac{df(u)}{du}\cdot \frac{du(t)}{dt}$ (7)

Let be $f(x) = (3 + \tan x)\cdot (4 + 3\cdot \cos x)$, then the derivative of $f(x)$ is:

$f'(x) = \sec^{2}x\cdot (4 + 3\cdot \cos x) + (3+\tan x)\cdot (-3\cdot \sin x)$

$f'(x) = 4\cdot \sec^{2}x+3\cdot \sec x -9\cdot \sin x -3\cdot \sin x\cdot \tan x$

$f'(x) = \frac{4+3\cdot \cos x}{\cos^{2}x} -\frac{9\cdot \sin x \cdot \cos x+3\cdot \sin^{2}x}{\cos x}$

$f'(x) = \frac{4+3\cdot \cos x-9\cdot \sin x \cdot \cos^{2}x+3\cdot \sin^{2}x\cdot \cos x}{\cos^{2}x}$

The derivative of $f(x) = (3 + \tan x)\cdot (4 + 3\cdot \cos x)$ is $f'(x) = \frac{4+3\cdot \cos x-9\cdot \sin x \cdot \cos^{2}x+3\cdot \sin^{2}x\cdot \cos x}{\cos^{2}x}$.

We kindly invite to check this question on derivatives: https://brainly.com/question/21202620