Question

y=(x/sqrt.(a^2 - 1)) - (2/sqrt.(a^2 - 1)) * arctan(sinx/(a+ sqrt.(a^2 - 1) + cosx)))

Answer 1

really, thats intresting

Answer 2

prove that y' = 1/(a+cosx)

Answer 3

\[y=(\frac{x}{\sqrt{a^2-1}}-\frac{2}{\sqrt{a^2-1}}) \cdot \tan^{-1}(\frac{\sin(x)}{a+\sqrt{a^2-1}}+\cos(x))\]?

Answer 4

better format: \[y=\left(\begin{matrix}x \\ \sqrt{a ^{2}-1} \end{matrix}\right) - \left(\begin{matrix}2 \\ \sqrt{a^2 - 1}\end{matrix}\right)\arctan(\left(\begin{matrix}sinx \\ a + \sqrt{a^2+1} + cosx\end{matrix}\right)\]

Answer 5

wait is my arctan part right are yours?

Answer 6

or yours?*

Answer 7

\[y=\frac{x}{\sqrt{a^2-1}}-\frac{2}{\sqrt{a^2-1}} \cdot \tan^{-1}(\frac{\sin(x)}{a+\sqrt{a^2-1}}+\cos(x))\]

Answer 8

mine, the cosx should be under the sinx

Answer 9

that's ugly ;)

Answer 10

\[(\frac{x}{\sqrt{a^2-1}})'=\frac{1}{\sqrt{a^2-1}}(x)'=\frac{1}{\sqrt{a^2-1}}\]
now we need to do the hard part find
\[(-\frac{2}{\sqrt{a^2-1}} \tan^{-1}(\frac{\sin(x)}{a+\sqrt{a^2-1}+\cos(x)}))'\]
\[-\frac{2}{\sqrt{a^2-1}} (\tan^{-1}(\frac{\sin(x)}{a+\sqrt{a^2-1}+\cos(x)}))'\]
so to find the derivative of this term first I will say
LET
\[g=\tan^{-1}(\frac{\sin(x)}{a+\sqrt{a^2-1}+\cos(x)})\]
now take tan( ) of both sides
\[\tan(g)=\frac{\sin(x)}{a+\sqrt{a^2-1}+\cos(x)} (=\frac{opp}{adj})\]

so we need to find the hyp by using Pythagorean thm
\[hyp=\sqrt{\sin^2(x)+(a+\sqrt{a^2-1}+\cos(x))^2}\]
so anyways back to
\[\tan(g)=\frac{\sin(x)}{a+\sqrt{a^2-1}+\cos(x)} \]
we need to take derivative of both sides
\[\sec^2(g) g'=\frac{\cos(x)(a+\sqrt{a^2-1}+\cos(x))-\sin(x)(0+0+\cos(x))}
{(a+\sqrt{a^2-1}+\cos(x))^2}\]
\[\sec^2(g) g'=\frac{\cos(x)[a+\sqrt{a^2-1}+\cos(x)-\sin(x)]}{(a+\sqrt{a^2-1}+\cos(x))^2}\]
now to find g' we divide both sides by sec^2(g)
\[g'=\frac{\cos(x)[a+\sqrt{a^2-1}+\cos(x)-\sin(x)]}{(\sec^2(g))(a+\sqrt{a^2-1}+\cos(x))^2}\]
we can use our triangle to write sec^2(g) in terms of x
\[g'=\frac{\cos(x)[a+\sqrt{a^2-1}+\cos(x)-\sin(x)]}{(\frac{(a+\sqrt{a^2-1}+\cos(x))}{\sqrt{\sin^2(x)+(a+\sqrt{a^2-1}+\cos(x))^2}})^2(a+\sqrt{a^2-1}+\cos(x))^2}\]