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How do you determine dy/dx given #sqrtx+sqrty=25#?

Answer 1

Christopher Agresta

#dy/dx=-sqrt(y/x)#.

#sqrtx+sqrty=25#.

#:. d/dx(sqrtx+sqrty)=d/dx25=0#.

#:. d/dx(sqrtx)+d/dx(sqrty)=0#.

#:. 1/(2sqrtx)+d/dy(sqrty)*dy/dx=0............".[Chain Rule]"#.

#:. 1/(2sqrtx)+1/(2sqrty)*dy/dx=0#

#:. dy/dx=-1/(2sqrtx)/(1/(2sqrty))#

#:. dy/dx=-sqrt(y/x)#.

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Answer 2

Isabella Ackerman

I found: #(dy)/(dx)=-sqrt(y/x)#

Here you need to use implicit differentiation because your #y# is difficult to be left alone on one side as in our usual functons; but it is not a problem, only remember that #y# is itself a function of #x# and when you differentiate it you need to include the term #(dy)/(dx)# to take into account this dependence; the rest is as usual.

So let us differentiate:

#1/(2sqrt(x))+1/(2sqrt(y))(dy)/(dx)=0#

you see the appearance of the term #(dy)/(dx)# after differentiating #sqrt(y)#!!!

rearrange:

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

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Answer 3

Christopher Agresta

To determine dy/dx, you can differentiate both sides of the equation implicitly with respect to x, then solve for dy/dx.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.