Question

If x=a cos^3x and y=b sin^3x.find dy/dx?

Answer 1

`y' = -b/a tanx`

Explanation:

We have that:

`x/(acosx) = cos^2x`

and:

`y/(bsinx) = sin^2x`

Then:

`x/(acosx)+y/(bsinx) = 1`

Differentiate implicitly:

`(acosx+axsinx)/(a^2cos^2x) + (by'sinx -bycosx)/(b^2sin^2x) = 0`

`1/(acosx)+(xsinx)/(acos^2x) + (y')/(bsinx) -(ycosx)/(bsin^2x) = 0`

`(y')/(bsinx) = (ycosx)/(bsin^2x) - (xsinx)/(acos^2x) -1/(acosx)`

`y' = (ycosx)/(sinx) - (bxsin^2x)/(acos^2x) -(bsinx)/(acosx)`

Substitute now `x` and `y` from the original equations:

`y' = (bsin^3xcosx)/(sinx) - (abcos^3xsin^2x)/(acos^2x) -(bsinx)/(acosx)`

and simplify:

`y' = cancel(bsin^2xcosx) - cancel(bcosxsin^2x) -(bsinx)/(acosx)`

Then:

`y' = -b/a tanx`