Calculus: Use implicit differentiation to find dy/dx if x = cosy. How is this solved?

Answer 1

If x = cosy, then y = cos^-1 x
y=cos^-1 x
y' = -1/√(1-x^2)

Answer 2

you're taking the by-product with appreciate to x => dy/dx Use implicit differentiation simply by fact the variable dont agree. Use the product rule one the left facet two times to distinguish and whenever you diff in terms of y(the var doesnt agree) incorporate dy/dx. Then set up the words so as that on the left you have each and every thing that includes (dy/dx) take it as an undemanding element so u can remedy for it. flow the term it is multiplying dy/dx to divide on the different facet of the equation. Voula permit me be responsive to in case you prefer a worked out answer

Answer 3

start with x = cosy
differentiate with respect to x , treat y as function of x so y = y(x)

you will get immediatly
1 = siny * dy/dx
( dx/dx = 1 , and dcosy/dx = -siny * dy/dx well as i said immediatly chain rule ... )

)

so -1/siny = dy/dx.

Answer 4

implicit differentiation:

x = cos ( y )

dx = - sin ( y ) dy

dy/dx = - sin ( y )

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If you want to express dy/dx in terms of x,
use the trig identity cos^2 (y) + sin^2 (y) = 1

So
sin^2 ( y ) = 1 - cos^2 ( y )

Substituting:

sin^2 ( y ) = 1 - x^2

-sin (y) = - sqr ( 1 - x^2 )

So

dy/dx = -sqr ( 1 - x^2)

Answer 5

this should be =====> (x)' = (cos y)'

(x)' = 1

(cos y)' = -sin(y) * (y)' as the chain rule says....

then dy/dx=1/-sin(y)= - csc (y)

Answer 6

x = cosy
1=dy/dx(-siny)
dy/dx=1/(-siny)
dy/dx=-1/√(1-x^2)