Certainly! Let's find the derivative [tex]\( f'(x) \)[/tex] and the value of the derivative at [tex]\( x = 3 \)[/tex] for the function [tex]\( f(x) = \frac{3x^2 \tan x}{\sec x} \)[/tex].
### Step-by-Step Solution
1. Simplify the Function [tex]\( f(x) \)[/tex]:
We start by simplifying the expression. Recall that [tex]\( \sec(x) = \frac{1}{\cos(x)} \)[/tex], hence:
[tex]\[
f(x) = \frac{3x^2 \tan x}{\sec x} = 3x^2 \tan(x) \cos(x)
\][/tex]
Here, we used the identity [tex]\( \sec(x) = \frac{1}{\cos(x)} \)[/tex] to simplify [tex]\( \frac{\tan(x)}{\sec(x)} \)[/tex] to [tex]\( \tan(x) \cos(x) \)[/tex].
2. Rewrite [tex]\( f(x) \)[/tex] Using Trigonometric Identities:
Recall that [tex]\( \tan(x) = \frac{\sin(x)}{\cos(x)} \)[/tex]. Therefore,
[tex]\[
f(x) = 3x^2 \left( \frac{\sin(x)}{\cos(x)} \right) \cos(x) = 3x^2 \sin(x)
\][/tex]
3. Differentiate [tex]\( f(x) \)[/tex] with Respect to [tex]\( x \)[/tex]:
To find [tex]\( f'(x) \)[/tex], we need to use the product rule. The product rule states that if [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex] are functions of [tex]\( x \)[/tex], then:
[tex]\[
\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
\][/tex]
Let [tex]\( u(x) = 3x^2 \)[/tex] and [tex]\( v(x) = \sin(x) \)[/tex]. Then:
[tex]\[
u'(x) = \frac{d}{dx}(3x^2) = 6x
\][/tex]
and
[tex]\[
v'(x) = \frac{d}{dx}(\sin(x)) = \cos(x)
\][/tex]
Applying the product rule:
[tex]\[
f'(x) = 6x \sin(x) + 3x^2 \cos(x)
\][/tex]
4. Substitute [tex]\( x = 3 \)[/tex] into [tex]\( f'(x) \)[/tex]:
To find [tex]\( f'(3) \)[/tex], substitute [tex]\( x = 3 \)[/tex] into the derivative:
[tex]\[
f'(3) = 6 \cdot 3 \cdot \sin(3) + 3 \cdot 3^2 \cdot \cos(3)
\][/tex]
Simplify the expression:
[tex]\[
f'(3) = 18 \sin(3) + 27 \cos(3)
\][/tex]
### Final Answers
[tex]\[
f'(x) = 6x \sin(x) + 3x^2 \cos(x)
\][/tex]
[tex]\[
f'(3) = 18 \sin(3) + 27 \cos(3)
\][/tex]
These are the first derivative of [tex]\( f(x) \)[/tex] and its value at [tex]\( x = 3 \)[/tex] respectively.
