How do you use implicit differentiation to find #(d^2y)/(dx^2)# given #5=4x^2+5y^2#?

Question

6496 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-04T20:41:22

See below.

#4x^2+5y^2=5#

Differentiate both sides of the equation.

#d/dx (4x^2+5y^2) = d/dx(5)#

#8x + 10 y dy/dx = 0#

I prefer to solve for #dy/dx# before differentiating again.

#dy/dx = (-4x)/(5y) = -4/5(x/y)#

Now differentiate again

#d/dx(dy/dx) = -4/5 d/dx(x/y)#

#(d^2y)/dx^2 = -4/5 ((1y-xdy/dx)/y^2)#

Now replace #dy/dx#

#(d^2y)/dx^2 = -4/5 ((1y-x (-4x)/(5y))/y^2)#

Simplify

# = -4/5((y + (4x)/(5y))/y^2)#

# = -4/5((((y + (4x)/(5y)) * 5y))/(y^2 * 5y))#

# = -4/5 ((5y^2+4x^2)/(5y^3)) #

And we're kind of done except that in the original statement of the question we're told that that numerator is #5#.

#4x^2+5y^2=5#, so we can simplify further:

#(d^2y)/dx^2 = - 4/(5y^3)#