Math Problem Statement

Differentiate the following using the derivative rule.

Solution

To differentiate each of these functions, we'll use various rules of differentiation such as the product rule, quotient rule, and trigonometric identities. Here are the derivatives:

  1. y=x2tanxy = \frac{x}{2 - \tan x}

    y=(2tanx)xsec2x(2tanx)2y' = \frac{(2 - \tan x) - x \sec^2 x}{(2 - \tan x)^2}

  2. y=sinθcosθy = \sin \theta \cos \theta

y=cos(2θ)y' = \cos(2\theta)

  1. f(θ)=secθ1+secθf(\theta) = \frac{\sec \theta}{1 + \sec \theta}

f(θ)=secθtanθ(1+secθ)sec2θ(1+secθ)2f'(\theta) = \frac{\sec \theta \tan \theta (1 + \sec \theta) - \sec^2 \theta}{(1 + \sec \theta)^2}

  1. y=cosx1sinxy = \frac{\cos x}{1 - \sin x}

y=(1sinx)(sinx)cos2x(1sinx)2y' = \frac{(1 - \sin x)(-\sin x) - \cos^2 x}{(1 - \sin x)^2}

  1. y=tsint1+ty = \frac{t \sin t}{1 + t}

y=(1+t)(sint+tcost)tsint(1+t)2y' = \frac{(1 + t)(\sin t + t \cos t) - t \sin t}{(1 + t)^2}

  1. y=1secxtanxy = \frac{1 - \sec x}{\tan x}

y=tanx(secxtanx)(1secx)sec2xtan2xy' = \frac{\tan x \cdot (-\sec x \tan x) - (1 - \sec x)\sec^2 x}{\tan^2 x}

  1. g(x)=6x22xg(x) = \frac{6x^2}{2 - x}

    g(x)=(2x)12x+6x2(2x)2g'(x) = \frac{(2 - x) \cdot 12x + 6x^2}{(2 - x)^2}

  2. R(w)=3w+w42w2+1R(w) = \frac{3w + w^4}{2w^2 + 1}

    R(w)=(2w2+1)(3+4w3)(3w+w4)4w(2w2+1)2R'(w) = \frac{(2w^2 + 1)(3 + 4w^3) - (3w + w^4) \cdot 4w}{(2w^2 + 1)^2}

  3. ( f(x) = \frac{\sqrt{x} + 2x}{7x -

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differentiation
Quotient Rule
Product Rule
Trigonometric Differentiation

Formulas

Quotient Rule: (u/v)' = (u'v - uv')/v^2
Product Rule: (uv)' = u'v + uv'
Trigonometric Derivatives: d/dx(sin x) = cos x, d/dx(cos x) = -sin x, d/dx(tan x) = sec^2 x, d/dx(sec x) = sec x tan x

Theorems

Fundamental rules of differentiation

Suitable Grade Level

Grades 11-12