Why is Implicit Differentiation Different from Explicit Differentiation?

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Implicit differentiation differs from explicit differentiation primarily due to the presence of both x and y variables in the equation, which prevents it from being expressed in the standard form of a function, f(x) = y. In explicit differentiation, the function is clearly defined, allowing for straightforward calculation of the derivative and subsequent tangent line equations. However, in implicit differentiation, the equation is often set to zero, complicating the process of isolating y and finding the slope. This complexity arises because the derivative involves both x and y terms, making it necessary to manipulate the equation to derive the correct values for x and y before determining the slope. Understanding these differences is crucial for solving problems involving implicit differentiation effectively.
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Homework Statement




Find the equations of the lines that pass through (0,0) and are tangent to x^2 - 4x + y^2 + 1 = 0


My confusion

I've been given a problem of this sort recently, except now it involves implicit differentiation. I know "how" to get to the correct answer. I just don't "understand" what's going on as I'm getting to the correct answer.

The problem I had recently was: Find the equations of the lines that pass through (0,0) and are tangent to "f(x) = x^3 - 9x^2 - 16x = y

When comparing these two problems, the process of finding the answer seems to be different, however, the question is essentially asking the same exact thing in both problems. So I must not be seeing "how" the process of solving each is actually similar or the same.

Here is the first one that I solved:

implicita.jpg



However, when I follow the same exact process for the current problem involving implicit differentiation, I get this:


implicitc.jpg




In the first problem, I am supposed to be looking for the equation of a line: (y-y)=m(x-x) and it becomes f(x) = (f'(x))(x)

However, the case of implicit differentiation does not seem to want to follow the same process, which leads me to the observation that:

in the first non implicit problem, f(x) = x^3 - 9x^2 - 16x fits neatly into the left side of equation y=f'x(x)

while

in the implicit problem f(x) = x^2 - 4x + y^2 + 1 = 0 does not fit neatly into the left side of equation y=f'x(x)

Why does f(x) = x^2 - 4x + y^2 + 1 = 0 not fit neatly into the left side of equation y = f'x(x) ? I thought that f(x) always equals y, since the equation for a function seems to always be expressed as "f(x) = y"
-----------------------------

Also, in the non implicit problem, f'(x) = 3x^2 - 18x - 16 seems to fit neatly into the m term of y=f'x(x)

while

in the implicit problem, f'(x) = 2x - 4 + 2y(dy/dx) = 0 does not fit neatly into the m term of y = f'x(x)

Why does f'(x) = 2x - 4 + 2y(dy/dx) = 0 not fit neatly in m? I thought that f'(x) was supposed to represent the slope of the tangent, thus fitting neatly into the m term (since it works that way in the non implicit problem).
------------------------------

Is it the fact that the equation is set equal to zero in the implicit problem while the function in the first problem is not set equal to zero? Does that have any impact on the process of solving this problem, if so, what impact does the zero have?

Is it the fact that there are x's and y's together in the implicit problem that is the factor that doesn't allow the entire f'x to fit neatly into the m term?

Why must we change the functions in order to arrive at the correct x value?
 
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You write, repeatedly, "f(x)" while there is no such "f". y is NOT a function of x.
 


HallsofIvy said:
You write, repeatedly, "f(x)" while there is no such "f". y is NOT a function of x.

This was a very confusing problem as it is. But the fact that I didn't understand that if an equation is in terms of both x and y's on one side, then it can't be considered a function of the form "f(x) = y. Thus its no wonder why it does fit so neatly. I would either have to get everything in terms of x before taking the derivative...or, as an easier route, i could do some substitution to solve for x. Once I solve for x, I then need to find y. Once I have both x and y values, I can then find the slope value. Then, the equation of the line(s) can be formed.

Thanks
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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