Implicit differentiation (example walkthrough) (video) | Khan Academy

Derivatives of inverse functions (video) | Khan Academy

Implicit Diffrentiation

Diffrentiation for functions that do not pass the vertical line test

1. We can write $y^3 + x^3 = 2xy$ as: $y(x)^3 + x^3 = 2xy(x)$
2. Then take the derivative with respect to $x$ of both sides: $\frac{d}{dx}(y(x)^3 + x^3) = \frac{d}{dx}(2xy(x))$
3. This will then simplify to: $3y^2y’ + 3x^2 = 2y+2xy’$
4. Now, sub in a point such as $(1,1)$ for $(x,y)$, which gives us: $3(1^2)y’ + 3(1)^2 = 2(1) = 2(1)y’$
5. $y’ = -1$, which is the slope of the tangent line to the original function

<aside> 💭 If the limit of $g(x)$ as $x \to a$ is equal to $M$, and $f(x)$ is continuous at $M$, then $$ the limit of $f(g(x))$ as $x \to a$ is equal to $$f

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