Question

I'm facing a problem with implicit differentiation can anyone help me with that ?

Answer 1

Sure. Whats the problem?

Answer 2

I can't find a specific rule to deal with it :(

Answer 3

i have an exam after 3 hours and i don't know what to do :(

Answer 4

you know how to do regular derivatives right?

Answer 5

yes i know

Answer 6

implicit differentiation makes use of the chain rule to differentiate implicitly defined functions.

here are two ways to define functions, implicitly and explicitly. Most of the equations we have dealt with have been explicit equations, such as y = 2x-3, so that we can write y = f(x) where f(x) = 2x-3. But the equation 2x-y = 3 describes the same function. This second equation is an implicit definition of y as a function of x. As there is no real distinction between the appearance of x or y in the second form, this equation is also an implicit definition of x as a function of y

Not all implicit equations can be restated explicitly in a single equation. For example, the implicit equation x^2+y^2 = 9 needs two explicit equations,

\[y=+\sqrt{9-x^2} and y=-\sqrt{9-x^2}\]

which are the top and bottom halves of a circle respectively, to define the functional relation completely.

And xy = sin(y)+x^2y^2 (the magenta curves in the figure at the left) cannot be solved for either y as an explicit function of x or x as an explicit function of y. This implicit function is considered in Example 2.
Perhaps surprisingly, we can take the derivative of implicit functions just as we take the derivative of explicit functions. We simply take the derivative of each side of the equation, remembering to treat the dependent variable as a function of the independent variable, apply the rules of differentiation, and solve for the derivative. Returning to our original example:

\[2x-y=-3\]
\[\frac{ d }{ dx }(2x-y)=0\]
\[2-\frac{ dy }{ dw }=0\]
\[\frac{ dy }{ dx }=2.\]

This is of course what we get from differentiating the explicit form, y = 2x-3, with respect to x.
This simple example may not be very enlightening. Consider the second example, the equation that describes a circle of radius 3, centered at the origin. Taking the derivative of both sides with respect to x, using the power rule for the derivative of y,

\[y^2+x^2=9\]
\[2y \frac{ dy }{ dx }+2x=0\]
\[\frac{ dy }{ dx }=-\frac{ x }{ y }.\]

It can be seen from the figures that for either part of the circle, the slope of the tangent line has the opposite sign of the ratio x/y, and that the magnitude of the slope becomes larger as the tangent point nears the x-axis.

(For this case, finding dy/dx as an explicit function of x requires use of the power rule for fractional powers, usually considered later. This example may be thought of as a taste of things to come.)

Answer 7

does that help or need clearer explanation

Answer 8

it's looks difficult to understand :( , thanks GirlByte but i think i need a clearer explanation

Answer 9

have you read it all

Answer 10

try to read. and understand

Answer 11

yes of course i read all of it

Answer 12

ok so tell me what are you actually doing when you are finding a simple derivative like y = x^2

Answer 13

simply it's y'=2x

Answer 14

also Sary,

\[\Huge{\color{teal}{\mathbb{\text{Welcome To Openstudy!}}}}\]

Answer 15

yes but teachers don't describe this whenever you are getting a derivative you are actually doing something like this \[y = x^2\] \[y' = \frac{ d }{ dx } (x^2)\] \[y' = 2x \frac{ dx }{ dx }\]

Answer 16

i read GirlByte's reply again and i have been understand a little bit of it

Answer 17

whenever you differentiante with respect to something that term goes with it. and because dx/dx = 1 the derivative of is 2x

Answer 18

now it's clear

Answer 19

\[\large{\color{turquoise}{\cal{\text{yay. If you have anymore questions, feel free to ask us.}}}}\]

Answer 20

if you wanna differentiate some function with respect to x, you get dy/dx if function is f(x) = y

Answer 21

thank you all ♥

Answer 22

\[\normalsize{\color{teal}{\mathbb{\text{No problem :)}}}}\]

Answer 23

and thus in @GirlByte examples she gets dy/dx while differentiating y