Finding the derivative of trig function

[josh_123]

Find dy/dx of these functions:

1. y=sin^-1(1/3x)
dy/dx=1/(√1-(1/3x)² (1/3)
dy/dx=1/(3√1-(1/9x²)

However the book answer is 1/√9-x²). How did they get this?

2. y=sin^-1(1/x)
dy/dx=1/√1-(1/x)² (-1/x²)
If i tried to simplified it by multiply it by x^2/x^2then
dy/dx= -x^2/x⁴√x²-1 which equal -1/x²√x²-1 However the book answer is -1/absx√x²-1. Can you explain to me how they get this?

3. find dy/dx by implicit differentiation
sin^-1(xy)=cos^-1(x-y)
so d/dx[sin^-1(xy)=cos^-1(x-y)]
1/√1-(xy)² (y+x) (dy/dx)=-1/√1-(x-y)² (0)
(y+x)/√1-(xy)^2 (dy/dx)=0
dy/dx= (- y+x)/(√1-(xy)²)

The formula i used for all these problems are
d/dx[sin^-1u]=1/√1-u² (du/dx)
d/dx[cos^-1u]=-1/√1-u² (du/dx)

~Thank you so much in advance. I really need help before my big test!

 

[QuarkCharmer]

For the first one, they just stuck the 3 in the denominator that came from the chain rule back into the radical.

\frac{1}{3\sqrt{1-\frac{1}{9}x^{2}}}
\frac{1}{\sqrt{9(1-\frac{1}{9}x^{2})}}
\frac{1}{\sqrt{(9-\frac{9}{9}x^{2})}}
\frac{1}{\sqrt{(9-x^{2})}}

 

[josh_123]

thank you what about the other two?