Math Problem Statement

Answer them and make them more simple and get straight to the point while still using limit theorem, lets start again from number 7

Determine the derivatives of the given functions (numbers 2-10) using differentiation rules.

Solution

Let's start by solving problem number 7 using the limit definition of derivatives. The function given is:

f(x)=cos(5x)tan(5x)f(x) = \cos(5x) \tan(5x)

Step 1: Apply the limit definition of the derivative

The derivative of f(x)f(x) is:

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Step 2: Use product rule for derivatives

We need to use the product rule because f(x)f(x) is a product of two functions, cos(5x)\cos(5x) and tan(5x)\tan(5x). The product rule states:

[ f'(x) = \frac{d}{dx} [\cos(5x) \cdot \tan(5x)] = \cos(5x) \cdot \frac{d}{dx}[\tan(5x)] + \tan(5x) \cdot \frac{d}{dx}[\cos(5x)] ]

Step 3: Derivatives of individual terms

  1. The derivative of tan(5x)\tan(5x) is 5sec2(5x)5 \sec^2(5x).
  2. The derivative of cos(5x)\cos(5x) is 5sin(5x)-5 \sin(5x).

Step 4: Substitute the derivatives back into the product rule

f(x)=cos(5x)5sec2(5x)+tan(5x)(5sin(5x))f'(x) = \cos(5x) \cdot 5 \sec^2(5x) + \tan(5x) \cdot (-5 \sin(5x))

Step 5: Simplify

  1. cos(5x)sec2(5x)=1cos(5x)cos(5x)=1\cos(5x) \cdot \sec^2(5x) = \frac{1}{\cos(5x)} \cdot \cos(5x) = 1, so the first term becomes 5sec2(5x)5 \sec^2(5x).
  2. The second term simplifies to 5sin(5x)tan(5x)=5sin(5x)sin(5x)cos(5x)=5sin2(5x)cos(5x)-5 \sin(5x) \cdot \tan(5x) = -5 \sin(5x) \cdot \frac{\sin(5x)}{\cos(5x)} = -5 \frac{\sin^2(5x)}{\cos(5x)}.

Thus, the final expression for the derivative is:

f(x)=5sec2(5x)5sin2(5x)cos(5x)f'(x) = 5 \sec^2(5x) - 5 \frac{\sin^2(5x)}{\cos(5x)}


Would you like more details or have any questions?

Here are 5 related questions to consider:

  1. How do you apply the limit definition of the derivative in simple functions?
  2. What is the product rule, and how does it simplify differentiating product functions?
  3. How do we differentiate trigonometric functions like sin(x)\sin(x) and tan(x)\tan(x)?
  4. Can you simplify trigonometric expressions involving sec(x)\sec(x) and sin(x)\sin(x)?
  5. How do you handle more complex functions in differentiation?

Tip: Always look for opportunities to apply the product and chain rules when differentiating complex expressions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differentiation
Trigonometric Functions
Product Rule
Chain Rule

Formulas

f'(x) = lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
Product Rule: (fg)' = f'g + fg'
Chain Rule: (f(g(x)))' = f'(g(x)) * g'(x)

Theorems

Limit Theorem
Product Rule
Chain Rule

Suitable Grade Level

Grade 11-12 (Advanced Calculus)