# help me with finding equation of tangent (implicit differentiation)

• Oct 17th 2009, 06:30 PM
zpwnchen
help me with finding equation of tangent (implicit differentiation)
• Oct 17th 2009, 07:16 PM
redsoxfan325
Quote:

Originally Posted by zpwnchen

Differentiate implicitly:

$6x^5+2x\frac{dy}{dx}+2y+4y^3\frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{6x^5+2y}{2x+4y^3}$

Plug in $(1,1)$ to find the slope and then use the point-slope equation to find the tangent line.
• Oct 17th 2009, 07:25 PM
zpwnchen
Why do i have to use implicit differentiation?
$
m = f_{x} (x,y)$
so $F_x = 6x^5+2y = 6+2 =8$.. why is not right?
• Oct 17th 2009, 07:31 PM
mr fantastic
Quote:

Originally Posted by zpwnchen
Why do i have to use implicit differentiation?
$
m = f_(x) (x,y)$
so $F_x = 6x^5+2y = 6+2 =8$.. why is not right?

.
This is wrong. You require the value of dy/dx at the given point. Note that y is an implicit function of x.
• Oct 17th 2009, 07:32 PM
redsoxfan325
Quote:

Originally Posted by zpwnchen
Why do i have to use implicit differentiation?
$
m = f_(x) (x,y)$
so $F_x = 6x^5+2y = 6+2 =8$.. why is not right?

This is not a function of three variables (i.e. $z=f(x,y)$) and therefore partial derivatives are not necessary because you are only trying to find $\frac{dy}{dx}$.
• Oct 17th 2009, 07:39 PM
zpwnchen
Quote:

Originally Posted by zpwnchen
Why do i have to use implicit differentiation?
$
m = f_{x} (x,y)$
so $F_x = 6x^5+2y = 6+2 =8$.. why is not right?

So this is right when y is not included in the function?
• Oct 17th 2009, 07:41 PM
zpwnchen
• Oct 17th 2009, 07:46 PM
mr fantastic
Quote:

Originally Posted by zpwnchen
So this is right when y is not included in the function?

Right for what? This makes no sense.

You are clearly confusing functions of several variables with functions of a single variable. To get the tangent to a curve defined by the relation f(x, y) = c you need to calculate dy/dx. To do this you either make y the subject (where possble or convenient) and differentiate or you just differentiate the relation using implicit differentiation and then solve for dy/dx. Either way, f(x, y) = c defines a function of a single variable.
• Oct 17th 2009, 07:56 PM
zpwnchen
i mean. if it's f(x) function(single variable) then i do not need to implicitly differentiate and just $F_{x}$ would give me m right?
Because it is a f(x y) function (several variable) so i have to implicit differentiate dy/dx to get m right?
• Oct 17th 2009, 08:05 PM
redsoxfan325
Quote:

Originally Posted by zpwnchen
i mean. if it's f(x) function(single variable) then i do not need to implicitly differentiate and just $F_{x}$ would give me m right?
Because it is a f(x y) function (several variable) so i have to implicit differentiate dy/dx to get m right?

You could think of the function as $x^6+2xf(x)+f^4(x)=4$.

That, you can see, is a function of one variable. If you could solve for $f(x)$, than you could just take a regular derivative, but since solving for $f(x)$ is either very hard or impossible, it makes the most sense to differentiate implicitly.