Question

PLEASE CALCULATE DERIVATIVE WITH RESPECT TO X: SIN^-1(A/X). (Comes from Problem Set 2 Section 5A.3F)

I don't understand how the chain rule generates the answer found in the solutions.

Answer 1

sin-1(a/x) = y ; im assuming "a" is a constant

a/x = sin(y)

-a/x^2 = y' cos(y) <-- by chain rule

a/(x^2 cos(y)) = y'
....................................

SO what is the answer to compare with?

Answer 2

sin(y) = opp/hyp = a/x

a^2 + adj^2 = x^2

adj = sqrt(x^2-a^2)

cos(y) = adj/hyp = sqrt(x^2 - a^2)/x

\[y'=-\frac{ax} {x^2\sqrt{x^2-a^2}}\]
\[y'=-\frac{a} {x\sqrt{x^2-a^2}}\]
does it look anything like that?

Answer 3

http://www.wolframalpha.com/input/?i=d%2Fdx+%28sin^-1%28a%2Fx%29%29

Answer 4

Thank you. Yes, that is the precisely the answer given in the solutions. I tried setting the equation to equal y and then doing implicit differentiation, but I didn't understand how to substitute for cosy- I forget that the Pythagorean theorem could be used in situations like this. Thanks also for the WA example, I think that is how the solutions derived it.

Could you now explain how d/du sin^-1= 1/sqrt (1-u^2) ?

Answer 5

im assuming its sin-1(u)=a this is simply definitional. the inverse sin of a ratio will tell us what angle it came from.

sin-1(u) = a
u/1 = sin(a) ; sin is defined as opposite/hypotenuse, so lets draw a picture.

Du(u) = Du(sin(a))

1 = cos(a) d/du

d/du = 1/cos(a) ; from the drawing, cos(a) = adj/hyp:
cos(a) = sqrt(1-u^2)

d/du = 1/sqrt(1-u^2)

Answer 6

something along those lines :)

Answer 7

Thank you again, very clearly explained. I would give another medal for this question if I could. In your own experience, how often is the Pythagorean Theorem used for solving derivatives? It think I need to be more proactive in exploiting it's use.

Answer 8

When dealing with the trig stuff; I find that it is always good to draw a picture so that you can visualize the process better.

It comes in handy when you have to do integration.