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Reginald Owens

Answered question

2022-02-28

My question is, how can I solve the following derivative question?

y = \sin (\arctan x) + \tan (\arcsin x)

Alexandra Haynes

Beginner2022-03-01Added 10 answers

First you want to simplify

\tan (\arcsin x)

and

\sin (\arctan x)

. Let me show you the latter one.
Suppose

θ = \arctan x

. Then by definition

- \frac{π}{2} < θ < \frac{π}{2}

and

\tan θ = x

. Since

\sec θ > 0

(because

- \frac{π}{2} < θ < \frac{π}{2}

), and

\sec^{2} θ = 1 + \tan^{2} θ = 1 + x^{2}

,
you get

\sec θ = \sqrt{1 + x^{2}}

, so

\cos θ = \frac{1}{\sqrt{1 + x^{2}}}

, and

\sin θ = \tan θ \cos θ = \frac{x}{\sqrt{1 + x^{2}}}

i.e.

\sin (\arctan x) = \frac{x}{\sqrt{1 + x^{2}}}

Similarly, you can prove

\tan (\arcsin x) = \frac{x}{\sqrt{1 - x^{2}}}

Using quotient rule, you can prove that the derivative of

\frac{x}{\sqrt{1 + x^{2}}}

and

\frac{x}{\sqrt{1 - x^{2}}}

are

\frac{1}{{(1 + x^{2})}^{\frac{3}{2}}}

and

\frac{1}{{(1 - x^{2})}^{\frac{3}{2}}}

respectively. Hence the derivative of your function is

\frac{1}{{(1 + x^{2})}^{\frac{3}{2}}} + \frac{1}{{(1 - x^{2})}^{\frac{3}{2}}}

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