Let y be defined implicitly by the equation ln(3y)=5xy Use implicit differentiation to find the first derivative of y with respect to x
a)dy/dx=
Use implicit differentiation to find the second derivative of y with respect to x
b)d2y/dx2=
c)Find the point on the curve where d2y/dx2=0
ln(3y) = 5xy
(a) Differentiating with respect to x,
(dy/dx)*1/y = 5(y+ xdy/dx)
By transposition we have,
dy/dx = (5y2)/(1 - 5xy) ....................(1)
(b) Again differentiating equation (1) with respect to x,
d2y/dx2 = { 10y dy/dx (1- 5xy) + 5(y+ xdy/dx)*(5y2) } / ( 1 - 5xy)2
On further solving we have,
d2y/dx2 = { 25y3 * (3 -10xy) } / ( 1- 5xy)3 ..........(2)
(c) For the required point we set equation (2) equal to 0,
Hence, d2y/dx2 = 0
y=0 or xy = 0.3
but y = 0 is not possible
So xy=0.3
substituting the value of xy in the curve we have
y = 1.493
x = 0.201
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