1 Let f(x) = (5 sin x + 8 cos a)tan f'(x) =

Calculus: Early Transcendentals
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Chapter1: Functions And Models
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Let \( f(x) = (5 \sin x + 8 \cos x) \tan^{-1} x \)

\[ f'(x) = \]

The task is to differentiate the function \( f(x) = (5 \sin x + 8 \cos x) \tan^{-1} x \). We will apply the product rule, which states that for two functions \( u(x) \) and \( v(x) \), the derivative is given by:

\[ (uv)' = u'v + uv' \]

Here, \( u(x) = 5 \sin x + 8 \cos x \) and \( v(x) = \tan^{-1} x \). 

1. Differentiate \( u(x) \):
   - \( u'(x) = 5 \cos x - 8 \sin x \)

2. Differentiate \( v(x) \):
   - \( v'(x) = \frac{1}{1+x^2} \)

Now substitute these into the product rule formula:

\[ f'(x) = (5 \cos x - 8 \sin x) \tan^{-1} x + (5 \sin x + 8 \cos x) \left( \frac{1}{1+x^2} \right) \]

This is the expression for the derivative \( f'(x) \).
Transcribed Image Text:Let \( f(x) = (5 \sin x + 8 \cos x) \tan^{-1} x \) \[ f'(x) = \] The task is to differentiate the function \( f(x) = (5 \sin x + 8 \cos x) \tan^{-1} x \). We will apply the product rule, which states that for two functions \( u(x) \) and \( v(x) \), the derivative is given by: \[ (uv)' = u'v + uv' \] Here, \( u(x) = 5 \sin x + 8 \cos x \) and \( v(x) = \tan^{-1} x \). 1. Differentiate \( u(x) \): - \( u'(x) = 5 \cos x - 8 \sin x \) 2. Differentiate \( v(x) \): - \( v'(x) = \frac{1}{1+x^2} \) Now substitute these into the product rule formula: \[ f'(x) = (5 \cos x - 8 \sin x) \tan^{-1} x + (5 \sin x + 8 \cos x) \left( \frac{1}{1+x^2} \right) \] This is the expression for the derivative \( f'(x) \).
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