`(x^2)y - 4x = 5` Find the second derivative implicitly in terms of `x` and `y`.

Textbook Question

Chapter 2, 2.5 - Problem 46 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`x^2y-4x=5`

Differentiating both sides with respect to x,

`x^2y'+y(2x)-4=0`

`x^2y'=4-2xy`

`y'=(4-2xy)/x^2`

Differentiating again both sides with respect to x,

`(d^2y)/dx^2=(x^2d/dx(4-2xy)-(4-2xy)d/dx(x^2))/x^4`

`(d^2y)/dx^2=(x^2(-2xdy/dx+y(-2))-(4-2xy)(2x))/x^4`

`(d^2y)/dx^2=(x^2(-2xdy/dx-2y)-2x(4-2xy))/x^4`

`(d^2y)/dx^2=(x(-2x^2y'-2xy-8+4xy))/x^4`

`(d^2y)/dx^2=(-2x^2y'+2xy-8)/x^3` 

`(d^2y)/dx^2=(-2(x^2y'-xy+4))/x^3`

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gsarora17 | (Level 2) Associate Educator

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In continuation of the answer, plug in the value of y' in y'',

`y''=(-2(4-2xy-xy+4))/x^3`

`y''=(-2(8-3xy))/x^3`

`y''=(6xy-16)/x^3`

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