How do you find #(d^2y)/(dx^2)# for #5=2x-5y^2#?

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Answer 1 · 2016-12-22T11:09:05

#(d^2y)/(dx^2) =-1/(sqrt5(5-2x)^(3/2))#

It would be possible to explicit #y(x)= +-sqrt((2x-5)/5)#, but we will use the procedure for implicit differentiation.

Differentiate the equation:

#2x-5y^2=5#

term by term with respect to #x#, considering that based on the chain rule:

#(dg(y))/(dx) = (dg(y))/(dy)*(dy)/(dx)#

So, we have:

#2-10y(dy)/(dx)= 0#

Solving for #(dy)/(dx)#:

#(dy)/(dx)= 1/(5y)#

Differentiating again:

#(d^2y)/(dx^2)= -1/(5y^2)(dy)/(dx) = -1/(5y^2)*1/(5y)=-1/(25y^3)#

Now from the original equation we have:

#5y^2=5-2x#

and we can use this to substitute #y# in the expression of the second derivative:

#(d^2y)/(dx^2) = -1/(25y^3) = -1/sqrt(5) * 1/(5y^2)^(3/2)=-1/(sqrt5(5-2x)^(3/2))#