How to get 2t2+1 as the derivative for sin⁡(θ) when tan⁡(θ2)=t

Using the Chain Rule directly,θ=arcsin⁡(2t1+t2)⇒dθdt=11−u2⋅2(1−t2(1+t2)2)[the part you found is only the dudt "Chain rule" factor]=11−(2t1+t2)2⋅(1−t2(1+t2)2)=1(1+t2)2−4t2(1+t2)2⋅2(1−t2(1+t2)2)=1(1−t2)2(1+t2)2⋅2(1−t2(1+t2)2)=(1+t21−t2)⋅2(1−t2(1+t2)2)=21+t2

"How to get 2t2+1 as the derivative for..." was answered by our community

Brielle James

Answered question

2022-04-06

How to get

\frac{2}{t^{2} + 1}

as the derivative for

\sin (θ)

when

\tan (\frac{θ}{2}) = t

Answer & Explanation

Kendall Wilkinson

Beginner2022-04-07Added 17 answers

Using the Chain Rule directly,

θ = \arcsin (\frac{2 t}{1 + t^{2}})

\Rightarrow \frac{d θ}{d t} = \frac{1}{\sqrt{1 - u^{2}}} \cdot 2 (\frac{1 - t^{2}}{{(1 + t^{2})}^{2}})

[the part you found is only the

\frac{d u}{d t}

"Chain rule" factor]

= \frac{1}{\sqrt{1 - {(\frac{2 t}{1 + t^{2}})}^{2}}} \cdot (\frac{1 - t^{2}}{{(1 + t^{2})}^{2}})

= \frac{1}{\sqrt{\frac{{(1 + t^{2})}^{2} - 4 t^{2}}{{(1 + t^{2})}^{2}}}} \cdot 2 (\frac{1 - t^{2}}{{(1 + t^{2})}^{2}})

= \frac{1}{\sqrt{\frac{{(1 - t^{2})}^{2}}{{(1 + t^{2})}^{2}}}} \cdot 2 (\frac{1 - t^{2}}{{(1 + t^{2})}^{2}})

= (\frac{1 + t^{2}}{1 - t^{2}}) \cdot 2 (\frac{1 - t^{2}}{{(1 + t^{2})}^{2}})

= \frac{2}{1 + t^{2}}

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