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### Implicit differentiation

Mar 25, 08:27 PM

It sometimes happens that a function y is defined by a simple implicit relation F(x,y)=0 while its explicit expression y=f(x) can by very complicated, or y cannot be expressed with elementary functions of x. In this case, to get information about the the derivative of the function y,we can use implicit differentiation.

For example if we consider the curve (strophoid) defined by

x(x²+y²)=2(x²-y²).

Question: What is the slope of the tangent line for

$\small \dpi{100}x=1,y= \sqrt{\frac{1}{3}}\text{ ?}$

Remark: All we need is

$\small \dpi{100} \frac{d y}{d x}(1)$,

written y'(1), to answer the problem. Expressing y as a function of x is not necessary. We will solve the problem using implicit differentiation.

Method: Think that y is a mysterious function of x, written y(x). We need its derivative at x=1. Let's try to solve the problem without finding an expression for y.

x(x²+y²(x)) is a function of x
2(X²-y(x)²) is a function of x too.

Those 2 functions are equal for any x therefore their derivatives are equal too. Let's differentiate both sides as functions of x. Of course, their derivatives equal

(x²+y²)+x(2x+2y'(x)y)=2(2x-2y'y)

Congratulation! We just calculated an implicit differentiation!

$\small \dpi{100}\text{At }x=1,y=\sqrt{ \frac{1}{3}}\text{, }$

$\small \dpi{100} \frac{4}{3}+2+\frac{2y'(1))}{\sqrt{3}}=4-\frac{4y'(1))}{\sqrt{3}}$

$\small \dpi{100} 6\frac{y'(1))}{\sqrt{3}}=4-2-\frac{4}{3}=\frac{2}{3}$

Answer:The slope of the tangent line at x=1 is $\small \dpi{100} y'(1)=\frac{\sqrt{3}}{9}$