Math Problem Statement

If f(x)=4sinx3+cosx\frac{4\sin x}{3+\cos x}, then f'(x)f′(x) =   

f'(3)f′(3) = 

Solution

Let's first compute the derivative of the function f(x)=4sinx3+cosxf(x) = \frac{4 \sin x}{3 + \cos x}.

Step 1: Apply the Quotient Rule

The quotient rule states that if we have a function of the form g(x)h(x)\frac{g(x)}{h(x)}, its derivative is:

(g(x)h(x))=g(x)h(x)g(x)h(x)(h(x))2\left( \frac{g(x)}{h(x)} \right)' = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}

Here, g(x)=4sinxg(x) = 4 \sin x and h(x)=3+cosxh(x) = 3 + \cos x.

  • Derivative of g(x)=4sinxg(x) = 4 \sin x: g(x)=4cosxg'(x) = 4 \cos x

  • Derivative of h(x)=3+cosxh(x) = 3 + \cos x: h(x)=sinxh'(x) = -\sin x

Step 2: Apply the Derivatives

Now, we apply the quotient rule:

f(x)=(4cosx)(3+cosx)(4sinx)(sinx)(3+cosx)2f'(x) = \frac{(4 \cos x)(3 + \cos x) - (4 \sin x)(-\sin x)}{(3 + \cos x)^2}

Simplify the numerator:

f(x)=4cosx(3+cosx)+4sin2x(3+cosx)2f'(x) = \frac{4 \cos x (3 + \cos x) + 4 \sin^2 x}{(3 + \cos x)^2}

Distribute 4cosx4 \cos x in the first term:

f(x)=12cosx+4cos2x+4sin2x(3+cosx)2f'(x) = \frac{12 \cos x + 4 \cos^2 x + 4 \sin^2 x}{(3 + \cos x)^2}

Step 3: Use the Trigonometric Identity

Recall that sin2x+cos2x=1\sin^2 x + \cos^2 x = 1, so:

f(x)=12cosx+4(1)(3+cosx)2f'(x) = \frac{12 \cos x + 4 (1)}{(3 + \cos x)^2}

Thus, the derivative simplifies to:

f(x)=12cosx+4(3+cosx)2f'(x) = \frac{12 \cos x + 4}{(3 + \cos x)^2}

Step 4: Evaluate f(3)f'(3)

Now, we compute f(3)f'(3).

First, find cos(3)\cos(3) and sin(3)\sin(3) (where 3 is in radians):

  • cos(3)0.989992\cos(3) \approx -0.989992
  • sin(3)0.14112\sin(3) \approx 0.14112

Substitute x=3x = 3 into the formula for f(x)f'(x):

f(3)=12(0.989992)+4(3+(0.989992))2f'(3) = \frac{12(-0.989992) + 4}{(3 + (-0.989992))^2}

f(3)=11.879904+4(30.989992)2f'(3) = \frac{-11.879904 + 4}{(3 - 0.989992)^2}

f(3)=7.879904(2.010008)2f'(3) = \frac{-7.879904}{(2.010008)^2}

f(3)=7.8799044.04016f'(3) = \frac{-7.879904}{4.04016}

f(3)1.9507f'(3) \approx -1.9507

Final Answers:

  • f(x)=12cosx+4(3+cosx)2f'(x) = \frac{12 \cos x + 4}{(3 + \cos x)^2}
  • f(3)1.9507f'(3) \approx -1.9507

Do you want further details or have any questions about the steps?
Here are 5 relative questions for you:

  1. How does the quotient rule work in other functions?
  2. What are the key trigonometric identities needed in calculus?
  3. How can you approximate trigonometric values for non-standard angles?
  4. What happens to f(x)f'(x) as xx approaches π\pi?
  5. Can you derive higher-order derivatives for the given function?

Tip: Always check for trigonometric identities when simplifying complex expressions involving sine and cosine.

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Math Problem Analysis

Mathematical Concepts

Calculus
Trigonometry
Derivatives
Quotient Rule

Formulas

Quotient rule: \( \left( \frac{g(x)}{h(x)} \right)' = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \)
Trigonometric identity: \( \sin^2 x + \cos^2 x = 1 \)

Theorems

Quotient Rule
Trigonometric Identities

Suitable Grade Level

Grades 11-12, Early College