Math Problem Statement

1. Find dy/dx for the given functions. 2. Find dv/dx for the given functions.

Solution

Let's go through the problems in the image one by one.

1. Find dydx\frac{dy}{dx}

  1. y=cos3(4x)y = \cos^3(4x)

    • Differentiate using the chain rule: dydx=3cos2(4x)(sin(4x))4=12cos2(4x)sin(4x)\frac{dy}{dx} = 3\cos^2(4x) \cdot (-\sin(4x)) \cdot 4 = -12 \cos^2(4x) \sin(4x)
  2. y=(sin(x)+1)4y = \left( \sin(x) + 1 \right)^4

    • Apply the chain rule: dydx=4(sin(x)+1)3cos(x)\frac{dy}{dx} = 4\left( \sin(x) + 1 \right)^3 \cdot \cos(x)
  3. y=7tan3(12x)y = 7 \tan^3(1 - 2x)

    • Differentiate using the chain rule: dydx=73tan2(12x)(2)sec2(12x)\frac{dy}{dx} = 7 \cdot 3 \tan^2(1 - 2x) \cdot (-2) \sec^2(1 - 2x) Simplifying: dydx=42tan2(12x)sec2(12x)\frac{dy}{dx} = -42 \tan^2(1 - 2x) \sec^2(1 - 2x)
  4. y=sin3(4x3)y = \sin^3(4x^3)

    • Differentiate using the chain rule: dydx=3sin2(4x3)cos(4x3)12x2=36x2sin2(4x3)cos(4x3)\frac{dy}{dx} = 3 \sin^2(4x^3) \cdot \cos(4x^3) \cdot 12x^2 = 36x^2 \sin^2(4x^3) \cos(4x^3)
  5. y=csc2(3x)y = \csc^2(3x)

    • Use the chain rule: dydx=2csc2(3x)(cot(3x))3=6csc2(3x)cot(3x)\frac{dy}{dx} = 2 \csc^2(3x) \cdot (-\cot(3x)) \cdot 3 = -6 \csc^2(3x) \cot(3x)
  6. y=(1cos(3x))3y = (1 - \cos(3x))^3

    • Use the chain rule: dydx=3(1cos(3x))2sin(3x)3=9(1cos(3x))2sin(3x)\frac{dy}{dx} = 3(1 - \cos(3x))^2 \cdot \sin(3x) \cdot 3 = 9(1 - \cos(3x))^2 \sin(3x)
  7. y=3sec3(2x)y = -3 \sec^3(2x)

    • Differentiate using the chain rule: dydx=33sec2(2x)sec(2x)tan(2x)2=18sec3(2x)tan(2x)\frac{dy}{dx} = -3 \cdot 3 \sec^2(2x) \cdot \sec(2x) \cdot \tan(2x) \cdot 2 = -18 \sec^3(2x) \tan(2x)
  8. y=2cos5(3x)y = -2 \cos^5(3x)

    • Use the chain rule: dydx=25cos4(3x)(sin(3x))3=30cos4(3x)sin(3x)\frac{dy}{dx} = -2 \cdot 5 \cos^4(3x) \cdot (-\sin(3x)) \cdot 3 = 30 \cos^4(3x) \sin(3x)
  9. y=(1tan2(x))y = (1 - \tan^2(x))

    • This simplifies using the identity 1tan2(x)=sec2(x)1 - \tan^2(x) = \sec^2(x), so: dydx=2sec2(x)sec2(x)=2sec4(x)\frac{dy}{dx} = 2 \sec^2(x) \cdot \sec^2(x) = 2 \sec^4(x)

2. Find dvdx\frac{dv}{dx}

  1. v=sin(3x)cos(2x)v = \sin(3x) \cos(2x)

    • Use the product rule and trigonometric identities: dvdx=cos(3x)3cos(2x)sin(3x)2sin(2x)\frac{dv}{dx} = \cos(3x) \cdot 3 \cos(2x) - \sin(3x) \cdot 2 \sin(2x)
  2. v=sin3(2x+1)v = \sin^3(2x + 1)

    • Use the chain rule: dvdx=3sin2(2x+1)cos(2x+1)2=6sin2(2x+1)cos(2x+1)\frac{dv}{dx} = 3 \sin^2(2x + 1) \cdot \cos(2x + 1) \cdot 2 = 6 \sin^2(2x + 1) \cos(2x + 1)
  3. v=2xcos(3x)v = 2x \cos(3x)

    • Use the product rule: dvdx=2cos(3x)2x3sin(3x)=2cos(3x)6xsin(3x)\frac{dv}{dx} = 2 \cos(3x) - 2x \cdot 3 \sin(3x) = 2 \cos(3x) - 6x \sin(3x)
  4. v=1tan(3x)v = \sqrt{1 - \tan(3x)}

    • Use the chain rule: dvdx=121tan(3x)(sec2(3x))3=3sec2(3x)21tan(3x)\frac{dv}{dx} = \frac{1}{2\sqrt{1 - \tan(3x)}} \cdot (-\sec^2(3x)) \cdot 3 = \frac{-3 \sec^2(3x)}{2\sqrt{1 - \tan(3x)}}
  5. v=3+2cos(2x)v = \sqrt{3 + 2 \cos(2x)}

    • Use the chain rule: dvdx=123+2cos(2x)(2sin(2x))2=2sin(2x)3+2cos(2x)\frac{dv}{dx} = \frac{1}{2\sqrt{3 + 2 \cos(2x)}} \cdot (-2 \sin(2x)) \cdot 2 = \frac{-2 \sin(2x)}{\sqrt{3 + 2 \cos(2x)}}

Would you like further explanations on any of these derivatives? Here are 5 related questions for deeper understanding:

  1. How can the chain rule be extended to multivariable functions?
  2. What is the product rule and when is it applied?
  3. How does the derivative of tan(x)\tan(x) result in sec2(x)\sec^2(x)?
  4. Why is ddxcsc(x)=csc(x)cot(x)\frac{d}{dx} \csc(x) = -\csc(x) \cot(x)?
  5. How do you apply trigonometric identities to simplify complex derivatives?

Tip: Always verify the derivative of trigonometric functions with identities before solving, as it often simplifies the work.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differentiation
Trigonometry
Chain Rule
Product Rule

Formulas

Chain rule: d/dx [f(g(x))] = f'(g(x)) * g'(x)
Product rule: d/dx [u(x) * v(x)] = u'(x) * v(x) + u(x) * v'(x)
Basic trigonometric derivatives: d/dx [sin(x)] = cos(x), d/dx [cos(x)] = -sin(x), d/dx [tan(x)] = sec^2(x)

Theorems

Chain Rule
Product Rule

Suitable Grade Level

Grades 11-12