Differential

Calculus Level 4

Let x = ln ⁡ tan ⁡ θ 2 x= \ln \tan \dfrac \theta2 and given that y = tan ⁡ θ − θ y= \tan \theta - \theta . Determined d 2 y d x 2 \frac{{d}^{2}y}{dx^{2}} if can be expressed in the form A tan ⁡ B θ . C sin ⁡ D θ ( E sec ⁡ F G θ + H ) A \tan B\theta . C \sin D\theta(E \sec^{F} G\theta+H) . Compute A + B + C + D + F + G + H A+B+C+D+F+G+H .


The answer is 10.

Muhammad Maulana by Muhammad Maulana , 25, Indonesia

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1 solution

Muhammad Maulana
Feb 6, 2016
  • Let x = ln ⁡ tan ⁡ θ 2 x= \ln \tan \dfrac \theta2
  • d x d θ = sec ⁡ 2 θ 2 2 tan ⁡ θ 2 \dfrac{dx}{d\theta} = \dfrac{\sec^{2}\dfrac\theta2} {2\tan \dfrac \theta2}
  • d x d θ = 1 2 sin ⁡ θ 2 x cos ⁡ θ 2 \dfrac{dx}{d\theta} = \dfrac 1{2\sin \dfrac \theta2 x \cos \dfrac \theta2}
  • d x d θ = 1 sin ⁡ θ \dfrac{dx}{d\theta} = \dfrac 1{\sin\theta}

  • Thus d θ d x = sin ⁡ θ \dfrac{d\theta}{dx} = \sin\theta ... ( 1 ) (1)

  • Then, let y = tan ⁡ θ − θ y= \tan \theta - \theta

  • d y d θ = sec ⁡ 2 θ − 1 \dfrac{dy}{d\theta} = \sec^{2}\theta-1
  • d y d θ = tan ⁡ 2 θ \dfrac{dy}{d\theta} = \tan^{2}\theta ... ( 2 ) (2)

Therefore d 2 y d x 2 = d ( t a n 2 θ . sin ⁡ θ ) d x \frac {d^{2}y}{dx^{2}} = \frac {d {(tan^{2}\theta . \sin\theta)}}{dx} => d 2 y d x 2 = 2 tan ⁡ θ . sec ⁡ 2 θ . sin ⁡ θ + tan ⁡ 2 θ . cos ⁡ θ \frac {d^{2}y}{dx^{2}} = 2 \tan \theta . \sec^{2} \theta . \sin \theta + \tan^{2} \theta . \cos \theta = tan ⁡ θ . sin ⁡ θ ( 2 sec ⁡ 2 θ + 1 ) \tan \theta . \sin \theta (2 \sec^{2} \theta +1) ∴ A + B + C + D + E + F + G + H = 1 + 1 + 1 + 1 + 2 + 2 + 1 + 1 = 10 \therefore A+B+C+D+E+F+G+H = 1+1+1+1+2+2+1+1 = \boxed{10}

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