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

The derivative of [tex]f(x) = (3 + \tan x)\cdot (4 + 3\cdot \cos x)[/tex] is [tex]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}[/tex].

In this question we shall use the following derivative rules:

Derivative of a sum of functions

[tex]\frac{d}{dt}(f(t) + g(t)) = \frac{df(t)}{dt} + \frac{dg(t)}{dt}[/tex] (1)

Derivative of a product of functions

[tex]\frac{d}{dt}(f(t)\cdot g(t)) = \frac{df(t)}{dt}\cdot g(t) + f(t)\cdot \frac{dg(t)}{dt}[/tex] (2)

Derivative of a constant

[tex]\frac{d}{dt}(k) = 0[/tex], [tex]\forall k \in \mathbb{R}[/tex] (3)

Derivative of a function multiplied by a constant

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

Derivative of the cosine function

[tex]\frac{d}{dt} \cos t = -\sin t[/tex] (5)

Derivative of the tangent function

[tex]\frac{d}{dt}\tan t = \sec^{2} t[/tex] (6)

Chain rule

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

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

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

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

[tex]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}[/tex]

[tex]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}[/tex]

The derivative of [tex]f(x) = (3 + \tan x)\cdot (4 + 3\cdot \cos x)[/tex] is [tex]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}[/tex].

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