Simple Implicit Differentiation

**JoeyCC** · November 20th 2008, 12:10 PM

Well, I live in Brazil and even though I'm not a complete wiz in math I have taken the dangerous path of Mathematics Higher Level at school. We follow the IB Curriculum. I have been struggling with calculas lately because I simply do not understand the concept of implicit differentiation. I was hoping someone could help me in a nice and digestible (very easy and slow) way to understand it. Please go slow on me cause I'm not the brightest bulb of the pack.

**JoeyCC** · November 20th 2008, 12:14 PM

Here's an example of what I mean:

Exercise 1. Find y' if xy3 + x2y2 + 3x2 - 6 = 1.

Answer. Let us use implicit differentiation. Then we take the derivative of the equation treating y as a function of x. We get

Algebraic manipulations give

I don't get where this second y' is coming from. I know the chain rule and product rule and I understand the rules of differenciation, however I suspect my lack of conceptual knowledge impedes me to understand where the y' comes from.

I know the rules of differentiation but I don't know what differentiation is actually doing. It's really come down to a mechanical process.

Please help me understand implict differentiation and what differentiation is doing.

~Kindest Regards

Joey

NOTE> Sorry for double posting

**elizsimca** · November 20th 2008, 08:44 PM

One thing to realize about implicit differentiation is that it is a nice tool to use when the equation is not "nice" to solve for in terms of y.

So let's consider the following... I am going to use $\frac{dy}{dx}$ instead of using $y'$ . They are the same thing...both of them mean "take the derivative of y with respect to x." I am going to show a very simplistic example..I will show you an equation that would be fine to solve for in terms of y, but hopefully it still gets the point across:

Okay, consider the equation $y=\frac{x^2+3x}{4}$

This is easy to differentiate, no need to implicitly differentiate because this equation is solved for already in terms of y...differentiating the normal way yields:

$\frac{dy}{dx}=\frac{d}{dx}(\frac{x^2+3x}{4})$

We can pull out that $\frac{1}{4}$ because it is a constant, so you end up with:

$\frac{dy}{dx}=(\frac{1}{4})\frac{d}{dx}(x^2+3x)=\f rac{1}{4}(2x+3)=\frac{x}{2}+\frac{3}{4}$

Now, let's suppose the equation was not written in terms of y and for some reason we would want to differentiate it implicitly:

Let's rewrite the equation like this:

Original equation: $y=\frac{x^2+3x}{4}$
Equation rewritten: $4y-3x=x^2$

So, implicitly differentiating our rewritten equation that is not solved for in terms of y would give us:

$4\frac{dy}{dx}-3=2x$

We need to get the terms (in this case only one term) with $\frac{dy}{dx}$ on one side of the equation, and all the terms without a $\frac{dy}{dx}$ on the other side. So we would need to add 3 to each side right? So the equation now becomes:

$4\frac{dy}{dx}=3+2x$

Now, to get $\frac{dy}{dx}$ by itself, we need to divide each side by 4. So we are solving for $\frac{dy}{dx}$ :

$\frac{dy}{dx}=\frac{3+2x}{4}=\frac{3}{4}+\frac{x}{ 2}=\frac{x}{2}+\frac{3}{4}$

Notice that this is exactly the answer we had when we differentiated directly when the equation was solved for y.

The main thing to remember about implicit differentiation is that x and y will be on both sides of the equation. Since we are differentiating y with respect to x, when we take the derivative of y we must have a $\frac{dy}{dx}$ attached to it, when we differentiate a term that only has an x, we don't need the $\frac{dy}{dx}$ because when we differentiate some term that has only an x in it, we are literally taking the derivative of that term with respect to x...and when you take the derivative of x with respect to x you don't need the differential hanging on to it.

Does this clear it up at all?

**tom@ballooncalculus** · November 21st 2008, 04:26 AM

Extra nice and digestible, I hope, is a diagram. So if the problem were, for example, OCR GCE Core Maths 4 (4724) 2007 question 6...

"The equation of a curve is $x^2 + 3xy + 4y^2 = 58$ . Find the equation of the normal at the point (2,3) on the curve, giving your answer in the form ax + by + c = 0, where a, b and c are integers."

... then we could draw this, to show the whole of the equation differentiated with respect x, before some 'normal' algebra i.e. collecting like terms and solving for dy/dx and using it as required...

Straight continuous lines differentiate downwards (or integrate upwards) with respect to x, while a straight dashed line differentiates (or integrates) with respect to the dashed balloon expression, so as to satisfy the chain rule. So we have 3 basic shapes...

Hope this helps - or doesn't further confuse! Balloon Calculus: worked examples from past papers

P.S. Only just noticed you posted your own example - will post the diagram later (if you haven't yourself!).

**JoeyCC** · November 21st 2008, 04:50 AM

I don't really get it but okay....I've noticed that that whenver you're implicit differentiating you always get dy/dx replacing the y.

Can someone find me a couple of examples to try? Let's see if I get it! Pick an easy one and a not so easy one please!

**tom@ballooncalculus** · November 21st 2008, 07:19 AM

Hi, JoeyCC!

Differentiation is called 'implicit' just because we can't immediately find $y$ or even its derivative totally in terms of $x$ . That's why the bottom row of the diagram has to include mentions of $y$ and $\frac{dy}{dx}$ (or $y'$ ).

But we differentiate the whole equation anyway, hoping to rearrange the resulting equation so that we have some sort of expression equal to $\frac{dy}{dx}$ .

Differentiating the whole equation means differentiating both sides, and each term in each side, with respect to the same variable, x. But if a term is a product we need the product rule, and if it's a composite function (one inside another) we need the chain rule.

y is a function of x, so any function of y is a composite.

So the derivative of $y^2$ with respect to $x$ is $2y$ times $\frac{dy}{dx}$ . (Chain rule.)

So try differentiating expressions like,

$y^4$

$xy$

$3x^3y^2$

all with respect to x, and then try again with the equations?

**JoeyCC** · November 21st 2008, 10:31 AM

I'm having difficulties formatting the text so I've put the contents of my post in a .doc

Please take the time to read it!

Sorry for the inconvenience.

**qorilla** · November 21st 2008, 12:44 PM

The following is the way I understand this thing. It may be more complicated for some, and more clear for some. I just want to show how I look at it.
This is basically the same as written above, but explained maybe deeper.

We are given a tough looking equation:

$x{y}^{2}+{x}^{2}{y}^{2}+3\,{x}^{2}-6={\rm e}^{xy}$

And are asked to find dy/dx.

If we want to find dy/dx, we actually consider y as a function of x, to make this clear, write:

y=f(x)

We suppose there is such an f(x) and it is differentiable, but we don't know what it actually is, neither its derivative.

Although from the above long equation we are not able to know what function f(x) is (infinitely many functions may be allowed by such an equation), we can still find it's derivative at least relative to f. I mean we can find the derivative in a form that uses the expression f(x) in it, this is sometimes the most exact result, since with complicated equations we don't know any formula for f(x), in that case we cannot expect finding f'(x) in a form that uses only x.

This is possible because we can treat the information we got by the equation the following way:

We know a function (let it be g(x)), written in two different closed forms (by two "formulas"), at least one of which uses our f(x) in it. (The two sides of the equation)
This function is in our case given in the following two forms:

$g \left( x \right) =x \left( f \left( x \right) \right) ^{2}+{x}^{2} \left( f \left( x \right) \right) ^{2}+3\,{x}^{2}-6 $
$ g \left( x \right) ={{\rm e}^{xf \left( x \right) }}$

Differentiate the two forms using the chain "rule" to get these monsters:

$ g'(x)=\left( f \left( x \right) \right) ^{2}+2\,xf \left( x \right) f' \left( x \right) +2\,x \left( f \left( x \right) \right) ^{2}+2\,{x}^{2}f \left( x \right) f' \left( x \right) +6\,x$
$ g'(x)={{\rm e}^{xf \left( x \right) }} \left( f \left( x \right) +xf' \left( x \right) \right)$

These forms must mean the same thing, so let's indicate this with an equals sign:

$\left( f \left( x \right) \right) ^{2}+2\,xf \left( x \right) f' \left( x \right) +2\,x \left( f \left( x \right) \right) ^{2}+2\,{x}^{2}f \left( x \right) f' \left( x \right) +6\,x = {{\rm e}^{xf \left( x \right) }} \left( f \left( x \right) +xf' \left( x \right) \right)$

Technical thing: Open up the multiplication on the right, move the terms containing f'(x) on one side, write it then in a form f'(x) * (...+...+...) = ...+...+..., divide by the ... on the left and you get something like (I was lazy and passed it to maple, I'm not sure it's correct, but the principle was the question):

$ f'(x)={\frac {{f(x)}^{2}+2\,x{f(x)}^{2}+6\,x-{{\rm e}^{xf(x)}}f(x)}{x \left( -2\,f(x)-2\, xf(x)+{{\rm e}^{xf(x)}} \right) }} $
Or

$\frac{dy}{dx}={\frac {{y}^{2}+2\,x{y}^{2}+6\,x-{{\rm e}^{xy}}y}{x \left( -2\,y-2\, xy+{{\rm e}^{xy}} \right) }}$

Now if someone says a pair of (x,y) [or (x,f(x)) ], we substitute the pair in the above formula to get dy/dx.

(We need to check first if the x,y pair at least fulfils our equation. It is the excercise's fault if it doesn't.)

**tom@ballooncalculus** · November 21st 2008, 02:11 PM

Hi JoeyCC - nearly there!

Your answers under the heading "Implicit differentiating" are correct (apart from one probable typo, see below), BECAUSE they're the derivatives of the given expressions with respect to x.

To go forward you need to appreciate that your answers under the heading "with respect to x" are no such thing.

$4y^3$ is the derivative of $y^4$ but with respect to y.

y + x would maybe be the result of applying the product rule to xy if this allowed you to differentiate y with respect to y and x with respect to x, but it has to be both with respect to the same variable. Which is what you've done under the other heading to get

$y+x\frac{dy}{dx}$

because it's

$\frac{dx}{dx}y+x\frac{dy}{dx}$

or

$x'y+xy'$

where $\frac{dx}{dx} = x' (with\ respect\ to\ x) = 1$ .

And exactly the same with the third example, except that you mean

${9x^2}{y^2} + {3x^3}{2y}{\frac{dy}{dx}}$

not

${9x^2}{y^2} + {3x^2}{2y}{\frac{dy}{dx}}$ .

By the way, for ${9x^2}{y^2} + {3x^3}{2y}{\frac{dy}{dx}}$ , the Latex code is

{9x^2}{y^2} + {3x^3}{2y}{\frac{dy}{dx}} enclosed by "math" in square brackets and "/math" in square brackets.

**JoeyCC** · November 21st 2008, 04:59 PM

Thank you man. You seem to know a lot about math
Plus can you please post one more example for me to try? If I get it right then we can move on.

~Kindest Regards

**tom@ballooncalculus** · November 22nd 2008, 01:27 AM

OK - differentiate some expressions with respect to x first, y second...

$\frac{d}{dx}{(x^2y)}$ is a useful notation to indicate the derivative of $x^2y$ with respect to x. So,

$\frac{d}{dy}{(x^2y)}$ means the derivative of $x^2y$ with respect to y.

$\frac{d}{dx}{(x^2y)} = 2xy + x^2y\frac{dy}{dx}$

$\frac{d}{dy}{(x^2y)} = 2x\frac{dx}{dy} * y + x^2 * 1 = 2xy\frac{dx}{dy} + x^2$

Do these, but make up your own examples too, for maximum confidence...

$\frac{d}{dx}{(y)} =$

$\frac{d}{dx}{(x)} =$

$\frac{d}{dy}{(y)} =$

$\frac{d}{dy}{(x)} =$

$\frac{d}{dx}{(x^4y)} =$

$\frac{d}{dy}{(x^4y)} =$

$\frac{d}{dx}{(3x^2y^2)} =$

$\frac{d}{dy}{(3x^2y^2)} =$

$\frac{d}{dx}{(3x^2siny)} =$

$\frac{d}{dy}{(3x^2siny)} =$

Do bear in mind though that in simple calculus we rarely need to do it both ways - this here is just because you seem to be confusing the two. Normally we make sure we know from the start which is the one variable of differentiation (assumed by the continuous lines in the diagrams), apart that is from the substitutions (often u or t or theta) that we bring in to represent the inner function for the chain rule (my dashed lines). This is true even for implicit differentiation - the first thing to decide for the given equation is whether you're looking for dy/dx (nearly always, and differentiate everything with respect to x) or dx/dy (everything w.r.t. y).

I would advise you to carry on using all the free help on this forum, which people are happy (as long as you aren't giving such forums a bad name by plagiarising, which you obviously aren't) to give for free. (In my case just to publicise the diagrams, and discover how they do or don't help.)

Balloon Calculus home page. Easy calculus! Don't integrate - balloontegrate!

**JoeyCC** · November 30th 2008, 03:53 PM

Yeah I don't get the difference between differentiating in respect to x and in respect to y. Can you please explain the difference using an example?

Thanks!

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I don't really get it but okay....

Here let's try....

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