Find the second derivative at the point (1,2), given the function below. y+3y^3-2x^2=10x+14
Mathematics · High School · Tue Nov 03 2020
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for finding the second derivative of the given function we wanna find the first derivative so
y+3y^3-2x^2=10x+14….(1)
By doing derivative of equation concerning ‘x’
Dy/Dx+3*3 *y^2(Dy/Dx)-2*2x=10 ……(2)
Dy/Dx(1+9y^2)-4x=10
Dy/Dx=(10+4x)/(1+9y^2) at (1,2)
Dy/Dx=(10+4*1)/(1+9(2)^2) at (1,2)
Dy/Dx=(14)/(37) at (1,2)
Doing again derivative of equation (2) concerning ‘x’
(D^2y/Dx^2)+9*2y (Dy/Dx) +9y^2(D^2y/Dx^2)-4=0
Putting the value of Dy/Dx at (1,2)
(D^2y/Dx^2)+36(14/37) +36(D^2y/Dx^2)-4=0
(D^2y/Dx^2)(37)=4-(36*14)/37
(D^2y/Dx^2)(37)=4-(36*14)/37
(D^2y/Dx^2)(37)=9.62
(D^2y/Dx^2)=9.62 /37
(D^2y/Dx^2)=13.5 answers.