Certainly! Let's solve the given problem step-by-step.
Given the function:
[tex]\[ f(x) = \frac{5 x^2 \tan x}{\sec x} \][/tex]
we want to find the derivative [tex]\( f'(x) \)[/tex] and then evaluate it at [tex]\( x = 3 \)[/tex].
1. Simplify the function:
We start by simplifying the function. Recall that [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex], which means:
[tex]\[ f(x) = \frac{5 x^2 \tan x}{\sec x} = 5 x^2 \tan(x) \cdot \cos(x) \][/tex]
Since [tex]\(\tan(x) = \frac{\sin(x)}{\cos(x)}\)[/tex], we can further simplify:
[tex]\[ f(x) = 5 x^2 \cdot \frac{\sin(x)}{\cos(x)} \cdot \cos(x) = 5 x^2 \sin(x) \][/tex]
2. Find the derivative [tex]\( f'(x) \)[/tex]:
Now, we need to differentiate [tex]\( 5 x^2 \sin(x) \)[/tex] with respect to [tex]\( x \)[/tex].
To find the derivative, we use the product rule for differentiation, which states that if you have a function [tex]\( h(x) = u(x) \cdot v(x) \)[/tex], then the derivative [tex]\( h'(x) \)[/tex] is given by:
[tex]\[ h'(x) = u'(x)v(x) + u(x)v'(x) \][/tex]
In our case, let:
[tex]\[ u(x) = 5 x^2 \][/tex]
[tex]\[ v(x) = \sin(x) \][/tex]
Find [tex]\( u'(x) \)[/tex] and [tex]\( v'(x) \)[/tex]:
[tex]\[ u'(x) = \frac{d}{dx}(5 x^2) = 10 x \][/tex]
[tex]\[ v'(x) = \frac{d}{dx}(\sin(x)) = \cos(x) \][/tex]
Now, apply the product rule:
[tex]\[ f'(x) = u'(x)v(x) + u(x)v'(x) \][/tex]
[tex]\[ f'(x) = (10 x) \sin(x) + (5 x^2) \cos(x) \][/tex]
[tex]\[ f'(x) = 10 x \sin(x) + 5 x^2 \cos(x) \][/tex]
3. Evaluate [tex]\( f'(3) \)[/tex]:
Substitute [tex]\( x = 3 \)[/tex] into the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(3) = 10 (3) \sin(3) + 5 (3)^2 \cos(3) \][/tex]
[tex]\[ f'(3) = 30 \sin(3) + 45 \cos(3) \][/tex]
Therefore, the simplified function's derivative is:
[tex]\[ f'(x) = 10 x \sin(x) + 5 x^2 \cos(x) \][/tex]
And the value of the derivative at [tex]\( x = 3 \)[/tex] is:
[tex]\[ f'(3) = 30 \sin(3) + 45 \cos(3) \][/tex]
Hence, [tex]\( f'(x) = 10 x \sin(x) + 5 x^2 \cos(x) \)[/tex] and [tex]\( f'(3) = 30 \sin(3) + 45 \cos(3) \)[/tex].
