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

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

please someone help me with this:
Consider the curve
The equation of the tangent line to the curve at the point has the form where
_________and _________________

2. Originally Posted by zpwnchen
please someone help me with this:

Consider the curve
The equation of the tangent line to the curve at the point has the form where
_________and _________________
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.

3. 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?

4. 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.

5. 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 $\frac{dy}{dx}$.

6. 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?

8. 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.

9. 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?

10. 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.