How do you find #(dy)/(dx)# given #sqrt(3x^7+y^2)=x#?

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Answer 1 · 2016-11-16T14:27:47

#sqrt((3x^7 + y^2))^2 = x^2#

#3x^7 + y^2 = x^2#

#21x^6 + 2y(dy/dx) = 2x#

#2y(dy/dx) = 2x - 21x^6#

#dy/dx =(2x- 21x^6)/(2y)#

However, since #x = 3x^7 + y^2#, we can substitute.

#dy/dx = (2sqrt(3x^7 + y^2) - 21x^6)/(2y)#

Finally, we must note our restrictions on the variable. The #√# can never be negative, so #3x^7 + y^2 ≥ 0#. Also, #y !=0#.

Hopefully this helps!