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Differentiation

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Question 17

By using the identities tan⁡θ=sin⁡θcos⁡θ\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}tanθ=cosθsinθ​ and sec⁡θ=1cos⁡θ\displaystyle \sec \theta = \frac{1}{\cos \theta}secθ=cosθ1​, and the standard derivatives of sin⁡θ \sin \theta\,sinθ and cos⁡θ\cos \thetacosθ, prove the following results:

a.
ddθ(tan⁡θ)=sec⁡2θ \frac{d}{d\theta}(\tan \theta) = \sec^2 \theta dθd​(tanθ)=sec2θ
[3]
b.
ddθ(sec⁡θ)=sec⁡θtan⁡θ \frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta dθd​(secθ)=secθtanθ
[3]
c.
ddθ(cot⁡θ)=−csc⁡2θ \frac{d}{d\theta}(\cot \theta) = -\csc^2 \theta dθd​(cotθ)=−csc2θ
[3]
d.
ddθ(csc⁡θ)=−csc⁡θcot⁡θ \frac{d}{d\theta}(\csc \theta) = -\csc \theta \cot \theta dθd​(cscθ)=−cscθcotθ
[3]

Differentiation Questions

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