Math Problem Statement

Suppose that f!\left(\frac{\pi}{4}\right)=-4 and f'!\left(\frac{\pi}{4}\right)=7, and let g(x) = f(x) \sin x and \displaystyle h(x) = \frac{\cos x}{f(x)}. Answer the following questions.

  1. Find g'(\pi/4). Answer: g'(\pi/4) =
  2. Find h'(\pi/4). Answer: h'(\pi/4) =

Solution

Let's solve both questions step by step.

1. Finding g(π/4)g'(\pi/4)

We are given:

  • g(x)=f(x)sinxg(x) = f(x) \sin x,
  • f(π4)=4f\left(\frac{\pi}{4}\right) = -4,
  • f(π4)=7f'\left(\frac{\pi}{4}\right) = 7.

To find g(x)g'(x), we use the product rule: g(x)=f(x)sinx+f(x)cosxg'(x) = f'(x) \sin x + f(x) \cos x Now, substitute x=π4x = \frac{\pi}{4}: g(π4)=f(π4)sin(π4)+f(π4)cos(π4)g'\left(\frac{\pi}{4}\right) = f'\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}\right) + f\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi}{4}\right) We know that: sin(π4)=cos(π4)=22\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} Substituting the values: g(π4)=722+(4)22g'\left(\frac{\pi}{4}\right) = 7 \cdot \frac{\sqrt{2}}{2} + (-4) \cdot \frac{\sqrt{2}}{2} Simplifying: g(π4)=722422g'\left(\frac{\pi}{4}\right) = \frac{7\sqrt{2}}{2} - \frac{4\sqrt{2}}{2} g(π4)=322g'\left(\frac{\pi}{4}\right) = \frac{3\sqrt{2}}{2}

2. Finding h(π/4)h'(\pi/4)

We are given:

  • h(x)=cosxf(x)h(x) = \frac{\cos x}{f(x)}.

To find h(x)h'(x), we use the quotient rule: h(x)=f(x)(sinx)cosxf(x)(f(x))2h'(x) = \frac{f(x)(-\sin x) - \cos x f'(x)}{(f(x))^2} Now, substitute x=π4x = \frac{\pi}{4}: h(π4)=f(π4)(sin(π4))cos(π4)f(π4)(f(π4))2h'\left(\frac{\pi}{4}\right) = \frac{f\left(\frac{\pi}{4}\right)(-\sin\left(\frac{\pi}{4}\right)) - \cos\left(\frac{\pi}{4}\right) f'\left(\frac{\pi}{4}\right)}{\left(f\left(\frac{\pi}{4}\right)\right)^2} Substituting the known values: h(π4)=422227(4)2h'\left(\frac{\pi}{4}\right) = \frac{-4 \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \cdot 7}{(-4)^2} h(π4)=2272216h'\left(\frac{\pi}{4}\right) = \frac{-2\sqrt{2} - \frac{7\sqrt{2}}{2}}{16} Combine the terms in the numerator: h(π4)=42272216=112216h'\left(\frac{\pi}{4}\right) = \frac{-\frac{4\sqrt{2}}{2} - \frac{7\sqrt{2}}{2}}{16} = \frac{-\frac{11\sqrt{2}}{2}}{16} Simplify: h(π4)=11232h'\left(\frac{\pi}{4}\right) = \frac{-11\sqrt{2}}{32}

Final Answers:

  1. g(π4)=322g'\left(\frac{\pi}{4}\right) = \frac{3\sqrt{2}}{2}
  2. h(π4)=11232h'\left(\frac{\pi}{4}\right) = \frac{-11\sqrt{2}}{32}

Would you like more details or clarifications on these solutions?

Here are 5 related questions to expand on this:

  1. How is the product rule applied to find derivatives of products of functions?
  2. What is the significance of the quotient rule in calculus?
  3. Can we interpret the physical meaning of g(x)g'(x) and h(x)h'(x) in certain contexts?
  4. How would the derivatives change if f(π4)f\left(\frac{\pi}{4}\right) or f(π4)f'\left(\frac{\pi}{4}\right) had different values?
  5. Can you apply implicit differentiation to find the derivatives of more complex functions?

Tip: Always check whether you are dealing with a product or quotient of functions before choosing the correct differentiation rule.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Product Rule
Quotient Rule
Trigonometric Functions

Formulas

Product rule: (uv)' = u'v + uv'
Quotient rule: (u/v)' = (vu' - uv') / v^2
Trig identities: sin(π/4) = cos(π/4) = √2/2

Theorems

Product Rule for Differentiation
Quotient Rule for Differentiation

Suitable Grade Level

Grades 11-12 (Advanced High School Calculus)