How does implicit differentiation work?

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How does implicit differentiation work?

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Answer 1 · 2014-08-05T09:41:47

Implicit differentiation is a way of differentiating when you have a function in terms of both x and y. For example:

$x^{2} + y^{2} = 16$ $x^2+y^2=16$

This is the formula for a circle with a centre at (0,0) and a radius of 4

So using normal differentiation rules $x^{2}$ $x^2$ and 16 are differentiable if we are differentiating with respect to x

$\frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) = \frac{d}{d x} (16)$ $d/dx(x^2)+d/dx(y^2)=d/dx(16)$

$2 x + \frac{d}{d x} (y^{2}) = 0$ $2x+d/dx(y^2)=0$

To find $\frac{d}{d x} (y^{2})$ $d/dx(y^2)$ we use the chain rule:

$\frac{d}{d x} = \frac{d}{d y} \cdot \frac{d y}{d x}$ $d/dx=d/dy *dy/dx$

$\frac{d}{d y} (y^{2}) = 2 y \cdot \frac{d y}{d x}$ $d/dy(y^2)=2y*dy/dx$

$2 x + 2 y \cdot \frac{d y}{d x} = 0$ $2x+2y*dy/dx=0$

Rearrange for $\frac{d y}{d x}$ $dy/dx$

$\frac{d y}{d x} = \frac{- 2 x}{2 y}$ $dy/dx=(-2x)/(2y$

$\frac{d y}{d x} = - \frac{x}{y}$ $dy/dx=-x/y$

So essentially to use implicit differentiation you treat y the same as an x and when you differentiate it you multiply be $\frac{d y}{d x}$ $dy/dx$

Youtube Implicit Differentiation

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How does implicit differentiation work?

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