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

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

answered by: Andrea S.
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