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Inverse Trig Derivatives
- Thread starter paulxzt
- Start date
Yes that is the answer I have but could you please be kind enough to show me how you got to that answer? Thanksgalactus said:\(\displaystyle \L\\\frac{d}{dx}\left[tan^{-1}(\frac{x}{a})\right]=\frac{a}{x^{2}+a^{2}}\)
\(\displaystyle \L\\y=tan^{-1}(\frac{x}{a})\)
and
\(\displaystyle \L\\\frac{x}{a}=tan(y)\)
\(\displaystyle \L\\\frac{d}{dx}\left[\frac{x}{a}\right]=sec^{2}(y)\frac{dy}{dx}\)
\(\displaystyle \L\\\frac{1}{a}=sec^{2}(y)\frac{dy}{dx}\)
\(\displaystyle \L\\\frac{1}{asec^{2}(\underbrace{y}_{\text{arctan(x/a)}})}=\frac{dy}{dx}\)
\(\displaystyle \L\\\frac{1}{asec^{2}(tan^{-1}(\frac{x}{a}))}=\frac{x^{2}+a^{2}}{a}\)
Therefore, \(\displaystyle \L\\=\frac{a}{x^{2}+a^{2}}\)
and
\(\displaystyle \L\\\frac{x}{a}=tan(y)\)
\(\displaystyle \L\\\frac{d}{dx}\left[\frac{x}{a}\right]=sec^{2}(y)\frac{dy}{dx}\)
\(\displaystyle \L\\\frac{1}{a}=sec^{2}(y)\frac{dy}{dx}\)
\(\displaystyle \L\\\frac{1}{asec^{2}(\underbrace{y}_{\text{arctan(x/a)}})}=\frac{dy}{dx}\)
\(\displaystyle \L\\\frac{1}{asec^{2}(tan^{-1}(\frac{x}{a}))}=\frac{x^{2}+a^{2}}{a}\)
Therefore, \(\displaystyle \L\\=\frac{a}{x^{2}+a^{2}}\)
ChaoticLlama
Junior Member
- Joined
- Dec 11, 2004
- Messages
- 199
it's no necessary to do it implicitly; it all depends on if you are more comfortable with trig or derivatives.
y = arctan(x/a)
differentiate, don't forget chain rule
dy/dx = (1 / (1 + x²/a²)) * 1/a
multiply by a/a to simplify fully
dy/dx = a / (a² + x²)
done
y = arctan(x/a)
differentiate, don't forget chain rule
dy/dx = (1 / (1 + x²/a²)) * 1/a
multiply by a/a to simplify fully
dy/dx = a / (a² + x²)
done