Lee Salinas
02/10/2024 · Middle School
If \( f(x)=\cos x-2 \tan x \), then \( f^{\prime}(x)=\square \) \( f^{\prime}(1)=\square \)
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Step-by-step Solution
To find the derivative of the function \( f(x) = \cos x - 2 \tan x \), we will use the chain rule and the derivatives of the trigonometric functions.
Given:
\[ f(x) = \cos x - 2 \tan x \]
To find \( f'(x) \), we will differentiate each term separately:
1. Differentiate \( \cos x \):
\[ \frac{d}{dx}(\cos x) = -\sin x \]
2. Differentiate \( -2 \tan x \):
\[ \frac{d}{dx}(-2 \tan x) = -2 \sec^2 x \]
Therefore, the derivative of \( f(x) \) is:
\[ f'(x) = -\sin x - 2 \sec^2 x \]
Now, to find \( f'(1) \), we substitute \( x = 1 \) into the derivative:
\[ f'(1) = -\sin(1) - 2 \sec^2(1) \]
Let's calculate the value of \( f'(1) \).
Calculate the value by following steps:
- step0: Calculate:
\(-\sin\left(1\right)-2\sec^{2}\left(1\right)\)
- step1: Calculate the trigonometric value:
\(-7.692509\)
The derivative of the function \( f(x) = \cos x - 2 \tan x \) is \( f'(x) = -\sin x - 2 \sec^2 x \).
To find \( f'(1) \), we substitute \( x = 1 \) into the derivative:
\[ f'(1) = -\sin(1) - 2 \sec^2(1) \]
Calculating the value of \( f'(1) \) gives us:
\[ f'(1) \approx -7.692509 \]
Quick Answer
\( f'(x) = -\sin x - 2 \sec^2 x \)
\( f'(1) \approx -7.692509 \)
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