How do you find #(dy)/(dx)# given #siny=x^2#?

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Answer 1 · 2017-01-24T17:52:22

#y'(x) = (2x)/sqrt(1-x^4)#

Differentiate the equation with respect to #x#:

#siny = x^2#

#y' cosy = 2x#

#y' = (2x) / cosy#

If we restrict the function to the intervals #x in [0,1]# and #y in [0,pi/2]#:

#cosy = sqrt( 1-sin^2y) = sqrt ( 1-x^4)#

so that:

#y'(x) = (2x)/sqrt(1-x^4)#