Answer:
f'(x) = 4x²sec(x)tan²(x) + 4x²sec^3(x) + 8xtan(x)sec(x): f'(1) ≈ 66.3766
Step-by-step explanation:
To find f'(x) of the function f(x) = 4x²tan(x)sec(x), we can use the product rule for three functions. This is given as:
[tex]\boxed{ \begin{array}{ccc} \text{\underline{Derivative of a Product of Three Functions:}} \\\\ \dfrac{d}{dx}[uvw]=uvw' + uv'w + u'vw \\\\ \text{Where:} \\ \bullet \ uvw \ \text{represents the product of functions } u, v, \text{ and } w \\ \bullet \ u', v', w' \ \text{represent the derivatives of } u, v, \text{ and } w \text{ with respect to } x \end{array}}[/tex]
Let's take the derivative:
[tex]\Longrightarrow \dfrac{d}{dx}[f(x) = 4x^2\tan(x)\sec(x)]\\\\\\\\\Longrightarrow \dfrac{d}{dx}[f(x)] = \dfrac{d}{dx}[4x^2\tan(x)\sec(x)][/tex]
[tex]\Longrightarrow f'(x)= 4x^2\tan(x)\dfrac{d}{dx}[\sec(x)] + 4x^2\dfrac{d}{dx}[\tan(x)]\sec(x) + \dfrac{d}{dx}[4x^2]\tan(x)\sec(x)[/tex]
To take the derivatives of the secant and tanget function we can use the following rules:
[tex]\boxed{\begin{array}{ccc}\text{\underline{Derivative of the Tangent Function:}}\\\\\dfrac{d}{dx}[\tan(x)] = \sec^2(x)\end{array}}[/tex]
[tex]\boxed{\begin{array}{ccc}\text{\underline{Derivative of the Secant Function:}}\\\\\dfrac{d}{dx}[\sec(x)] = \sec(x)\tan(x)\end{array}}[/tex]
We will also use the power rule:
[tex]\boxed{\begin{array}{ccc}\text{\underline{Power Rule for Differentiation:}}\\\\ \dfrac{d}{dx}[x^n]=nx^{n-1}\end{array}}[/tex]
Applying these rules, we get:
[tex]\Longrightarrow f'(x)= 4x^2\tan(x)(\sec(x)\tan(x)) + 4x^2(\sec^2(x))\sec(x) + (8x)\tan(x)\sec(x)[/tex]
Simplifying the above:
[tex]\therefore f'(x)= \boxed{4x^2\sec(x)\tan^2(x) + 4x^2\sec^3(x) + 8x\tan(x)\sec(x)}[/tex]
Now we are asked to evaluate the function when x = 1. Simply plug in 1 for 'x' and solve:
[tex]\Longrightarrow f'(1)= 4(1)^2\sec(1)\tan^2(1) + 4(1)^2\sec^3(1) + 8(1)\tan(1)\sec(1)\\\\\\\\\therefore f'(1) \approx \boxed{66.3766}[/tex]
Thus, all parts have been answered.
