How do you find #(dy)/(dx)# given #xsqrt(y+2)=4#?

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Answer 1 · 2017-03-09T20:53:03

#dy/dx = -(2(y+2))/x#

Differentiate both sides with respect to #x# using the product rule:

#d/dx (xsqrt(y+2)) = d/dx 4#

#sqrt(y+2) + x/(2sqrt(y+2))dy/dx = 0#

# + x/(2sqrt(y+2))dy/dx = -sqrt(y+2)#

#dy/dx = -(2(y+2))/x#

As #x > 0# we can also write it as:

#dy/dx = -(2xsqrt(y+2)sqrt(y+2))/x^2 = -(8*sqrt(y+2))/x^2#