Beck John
02/03/2023 · Senior High School
If \( f(x)=\frac{\tan x-4}{\sec x} \) \( f^{\prime}(x)=\square \) \( f^{\prime}(5)=\square \)
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Step-by-step Solution
To find the derivative \( f'(x) \) of the function \( f(x) = \frac{\tan x - 4}{\sec x} \), we can use the quotient rule. The quotient rule states that if \( f(x) = \frac{g(x)}{h(x)} \), then
\[
f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}
\]
In our case, let \( g(x) = \tan x - 4 \) and \( h(x) = \sec x \).
First, we need to find \( g'(x) \) and \( h'(x) \):
1. The derivative of \( g(x) = \tan x - 4 \) is:
\[
g'(x) = \sec^2 x
\]
2. The derivative of \( h(x) = \sec x \) is:
\[
h'(x) = \sec x \tan x
\]
Now we can apply the quotient rule:
\[
f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}
\]
Substituting \( g(x) \), \( g'(x) \), \( h(x) \), and \( h'(x) \):
\[
f'(x) = \frac{\sec^2 x \cdot \sec x - (\tan x - 4) \cdot \sec x \tan x}{(\sec x)^2}
\]
This simplifies to:
\[
f'(x) = \frac{\sec^3 x - (\tan x - 4) \sec x \tan x}{\sec^2 x}
\]
Now, we can simplify further:
\[
f'(x) = \frac{\sec^3 x}{\sec^2 x} - \frac{(\tan x - 4) \sec x \tan x}{\sec^2 x}
\]
This simplifies to:
\[
f'(x) = \sec x - \frac{(\tan x - 4) \tan x}{\sec x}
\]
Now, we can express \( \sec x \) in terms of \( \cos x \):
\[
f'(x) = \sec x - (\tan^2 x - 4 \tan x) \cos x
\]
Next, we need to evaluate \( f'(5) \). First, we need to calculate \( \tan(5) \) and \( \sec(5) \):
1. Calculate \( \tan(5) \) and \( \sec(5) \):
- \( \tan(5) \) is approximately \( -0.9589 \)
- \( \sec(5) = \frac{1}{\cos(5)} \) is approximately \( 1.1674 \)
Now substituting these values into \( f'(x) \):
\[
f'(5) = \sec(5) - \left( \tan^2(5) - 4 \tan(5) \right) \cos(5)
\]
Calculating \( \tan^2(5) \):
\[
\tan^2(5) \approx (-0.9589)^2 \approx 0.9195
\]
Now substituting into the expression:
\[
f'(5) = 1.1674 - \left( 0.9195 - 4(-0.9589) \right) \cos(5)
\]
Calculating \( \cos(5) \):
\[
\cos(5) \approx 0.2837
\]
Now substituting \( \cos(5) \):
\[
f'(5) = 1.1674 - \left( 0.9195 + 3.8356 \right) \cdot 0.2837
\]
Calculating the term in parentheses:
\[
0.9195 + 3.8356 \approx 4.7551
\]
Now substituting back:
\[
f'(5) = 1.1674 - 4.7551 \cdot 0.2837
\]
Calculating \( 4.7551 \cdot 0.2837 \):
\[
4.7551 \cdot 0.2837 \approx 1.3491
\]
Finally, we have:
\[
f'(5) = 1.1674 - 1.3491 \approx -0.1817
\]
Thus, the final answers are:
\[
f'(x
Quick Answer
\( f'(x) = \sec x - (\tan^2 x - 4 \tan x) \cos x \)
\( f'(5) \approx -0.1817 \)
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