Inverse Trig Q. (1 Viewer)

[collide]

Hi, can someone please help me find the derivative of:
y=sin(sin inverse x)

I did:
dy/dx = 1/sq rt.(1-x^2) cos (sin inverse x)

Apparently the answer is 1 according to the textbook.
Thanks!

 

[b0b444]

sure,

this is a bit of a trick question i guess...

do you remember that inverse functions cancel each other out? so if you take the inverse sine of something then take the sine of it, its like you done nothing to it.

therefore,

y = sin(sin inverse x)
= x

so dy/dx = 1

 

[morganforrest]

yeah it is just one... b0b444 got it right...the sine and inverse cancel each other out and y = x....hence dy/dx = 1

 

[collide]

oh i see.
thanks heaps for that

 

[SoulSearcher]

Hi, can someone please help me find the derivative of:
y=sin(sin inverse x)

I did:
dy/dx = 1/sq rt.(1-x^2) cos (sin inverse x)

Apparently the answer is 1 according to the textbook.
Thanks!

You could also get the same answer with this differentiation, if you figured out what cos (sin^-1 x) was without resorting to trigonometric functions. Note:
sin (sin^-1 x)
Let @ = sin^-1 x
Therefore x = sin @
Drawing the triangle, you find that cos @ = rt(1-x²)
But @ = sin^-1 x, therefore cos @ = cos (sin^-1 x) = rt(1-x²)

 

[collide]

You could also get the same answer with this differentiation, if you figured out what cos (sin^-1 x) was without resorting to trigonometric functions. Note:
sin (sin^-1 x)
Let @ = sin^-1 x
Therefore x = sin @
Drawing the triangle, you find that cos @ = rt(1-x²)
But @ = sin^-1 x, therefore cos @ = cos (sin^-1 x) = rt(1-x²)

That was really helpful. Thanks.

 

[SoulSearcher]

No problem