# For Implicit Differentiation What Exactly Does It Mean When It Says

Implicit differentiation--find tan(x/y)=x+y?

Ok when you differentiate y term with respect to x you differentiate with respect to y but you have to multiply that term by dy/dx
Chain rule is dy/dx=du/dx * dy/du
Quotient rule is if y=u/v then dy/dx=[v*du/dx - u*dv/dx]/v²

So let u=x/y then du/dx = [y*1 - x*dy/dx]/y²=[y-x*dy/dx]/y²
Now d/du (tan(u))=sec²(u) so hence d/dx (tan(x/y))=([y-x*dy/dx]/y²)*sec²(x/y)

So now differentiating both sides of the equation we get
([y-x*dy/dx]/y²)*sec²(x/y) = 1 + dy/dx
Now multiplying through by y² we get
(y-x*dy/dx)*sec²(x/y) = y² +y²*dy/dx

We now need to expand the brackets and collect all the dy/dx terms on one side and the other terms on the other side.
So we get
y*sec²(x/y) - y² =y²*dy/dx +x*sec²(x/y)*dy/dx
Now we factor out dy/dx to get
(y²+x*sec²(x/y))dy/dx=y*sec²(x/y) -y²
So now divide through by (y²+x*sec²(x/y)) to get dy/dx on its own so we get
dy/dx=[y*sec²(x/y) - y²]/(y²+x*sec²(x/y))
Hope this helps and good luck for Friday :)

What is implicit differentiation?

"Implicit differentiation" is a bit of a silly name  which just means differentiating both sides of an equation.  For instance, consider a curve given by the equation$(x^2+y^2)^3 = x^2 y^2$(Image courtesy WolframAlpha)Say we want to find the tangent line to this graph at a point on it.  We need to find $\frac{dy}{dx}$.  The straightforward way would be to try to break off a piece that's actually a function, solve for y, and then differentiate, but that's a lot of work.Instead, let's just take the derivative of both sides of the equation with respect to x.  This is easy to do using the chain rule and product rule:                  $3(x^2+y^2)^2\left(2 x + 2 y \frac{dy}{dx}\right) = 2 x^2 y \frac{dy}{dx} + 2 x y^2$Now just multiply this out and collect the dy/dx terms:      $6 x(x^2+y^2)^2 + 6 y(x^2+y^2)^2 \frac{dy}{dx}$  $= 2 x^2 y \frac{dy}{dx} + 2 x y^2$     $6 y(x^2+y^2)^2 \frac{dy}{dx} - 2 x^2 y \frac{dy}{dx}$  $= 2 x y^2 - 6 x(x^2+y^2)^2$And now we can just solve for dy/dx:                           $\frac{dy}{dx} = \frac{2 x y^2 - 6 x(x^2+y^2)^2 }{6 y(x^2+y^2)^2 -2 x^2 y}$ The term "implicit" is here because equations like $(x^2+y^2)^3 = x^2 y^2$ are said to give the relation between x and y "implicitly," as opposed to an equation like $y = \pm \sqrt{2 - x}$ which gives an "explicit" relation between x and y.  In the former equation, if we wanted to determine the y values where x = 0.1, we'd have to plug that in and then solve the equation $(y^2+0.01)^3 = 0.01 y^2$, whereas in the explicit relation we just have to plug it in and do the arithmetic.  So if we wanted to make any sense, we should call this technique "differentiating both sides of an implicit equation," not "implicit differentiation."  I prefer just to shorten that to "differentiating both sides," which conveys the idea just as well.

Why does implicit differentiation work?

Implicit differentiation is no different to many of the manipulations you already do with equations - as long as you do the same thing to both sides of an equality, then the equality is preserved. You’re very used to adding, subtracting, or scaling both sides of an equation. Differentiating (with respect to some variable) is just the same.Along the way though, you’ll run into a lot of familiar differentiation tactics, most notably the chain rule. Suppose one of your terms is a function of $y$, but you’re differentiating with respect to $x$. Not a problem, as long as you consider that locally $y$ can be treated as a function of $x$. But we don’t know explicitly what that function is. The best we can do is $d/dx\,f(y) = df/dy \cdot dy/dx$, so what we end up is another implicit equation, this time for $y’=dy/dx$, and then we will have to make $y’$ the subject.Applying operators throughout an equation will preserve the equality. Differentiation is just one, but there are many others that could be applied - generic differential operators, integration over an interval, limits, and so on. The most useful ones are, of course, linear operators - simply because we can do a term-by-term evaluation, and the equality persists, with the usual caveats with things such as infinities and limits.

Implicit Differentiation? xsin2y = ycos2x?

OK Cristina... implicit differentiation follows the same rules as regular differentiation. The only addition is that y is a function of x, and that means when to try to differentiate y with respect to x, or you try to differentiate a term with y in it, you need to use the chain rule for that term.

This is a good example of what you do.

xsin2y = ycos2x

Start by taking d / dx of both sides:
d[xsin2y]/dx = d[ycos2x]/dx

Now both of these terms have a y in it. We will use the product rule for both sides:
d[uv]/dx = v d[u]/dx + u d[dv]/dx
For the left-hand side u = x and v = sin2y; and for the right-hand side u = y and v = cos2x.

(sin2y) dx/dx + x d[sin2y]/dx = (cos2x) dy/dx + y d[cos2x]/dx

Now dx/dx = 1, and d[cos2x]/dx = - 2 sin2x using the chain rule. But we can also use the chain rule
on d[sin2y]/dx. Remember the chain rule says: if f is a function of u, and u is a function of x, then
df/dx = (df/du) (du/dx). In this case, f = sin2y, and y is a function of x.
So d[sin2y]/dx = (d[sin2y]/dy) dy/dx = (2cos2y) dy/dx.

Substituting what we determined back into the overall equation:
(sin2y) dx/dx + x d[sin2y]/dx = (cos2x) dy/dx + y d[cos2x]/dx

sin2y + x(2cos2y) dy/dx = (cos2x) dy/dx - 2y sin2x

Since we want to know what dy/dx is, we combine like terms and simplify:
(dy/dx) [2xcos2y - cos2x] = - 2y sin2x - sin2y
dy/dx = [- 2y sin2x - sin2y] / [2xcos2y - cos2x]

Implicit differentiation problem!?

Assuming you're differentiating with respect to x:

Remember whenever you differentiate y, you need to still have a dy/dx.

So:

5(x^2)(y^3) - 2(x^4) = -2/y

Don't forget to use the product rule with the first numbers.

=> [10x(y^3) + 15(x^2)(y^2)(dy/dx)] - 8(x^3) = 2(y^-2)(dy/dx)

So we bring all the (dy/dx) together:

=> 10x(y^3) -8(x^3) = 2(y^-2)(dy/dx) - 15(x^2)(y^2)(dy/dx)

Then distribute out the (dy/dx):

=> 10x(y^3) -8(x^3) = [2(y^-2) - 15(x^2)(y^2)](dy/dx)

Then divide by whatever is left:

=> [10x(y^3) -8(x^3)] / [2(y^-2) - 15(x^2)(y^2)] = dy/dx

And you're done!

Is this proof for implicit differentiation and explicit differentiation yielding the same result decent?

The proof is okay as far as it goes, but it's probably circular in the end.The question is, how do you know d/dx (lnx) = 1/x?Usually this is proved from the fact that d/dx (e^x) = e^x. Therefore, to go back and then use the fact about ln in a proof about e^x is circular.

How do I do implicit differentiation in Calculus?

Assume that $y$ is a function of $x$, or $y=f(x)$$y^2=f(x)^2$$\frac{d}{dx}y^2 = \frac{d}{dx}f(x)^2$Here is where the chain rule comes in to play:$\frac{d}{dx}f(x)^2=2 f(x) f^{\prime}(x)$                     $=2y^{\prime}y$                     $=2\frac{dy}{dx}y$$y$ is not defined explicitly as a function of $x$. That is why we must differentiate implicitly.  We also assume that $y$ is a function of $x$.  If you graph the original equation you will note that the graph can be split into pieces that are functions, but the equation as a whole is not a function of $x$.  That's why we use this technique.  Also, this is just a detailed explanation of what's going on.  I wouldn't recommend doing this for all problems, but in terms of understanding what you are doing this should clear some things up.  Hope this helps!Note: You would then solve for $\frac{dy}{dx}$ by isolating it just as you would solve for any variable.

In calculus we will be talking only about mathematical relationship that ARE functions (in Descartes' coordinates). If we will need to work with something that are not (like y = √x) we will specify what portion we want eliminate such that remaining portion of a graph represent a function (and will pass vertical line test).

However, there are plenty of math relationship that are a functions, however, you will not be able explicitly express y through the x (separate y from the x). In this case we assuming that y is still a function of x.

Example:

y³ + y² - 5y - x² = -4 we can differentiate explicitly, solving for y and then differentiate (which is very difficult), or do that implicitly:

d . . . . . . . . . . . . . . . d
--- [y³ + y² - 5y - x²] = ----[-4]
dx . . . . . . . . . . . . . . dx

. . . .dy . . . .dy . . .dy . . . . . . . . . .dy
3y² ---- + 2y ---- - 5 ---- - 2x = 0 ----> -----(3y² +2y - 5) = 2x Therefore:
. . . .dx . . . . dx . . .dx . . . . . . . . . .dx

dy . . . . . . .2x
---- = ----------------- (Much easier!)
dx . . . 3y² +2y - 5

In implicit differentiation why do you try to differentiate each side in respect to x?

Well typically we have an equation where y is the output.  y=f(x)f(x)^2+x^2=4We typically view x as the input, the independent variable, and y is the dependent variable.  It is dependent on the value of x.  We can rewrite y^2+x^2=4like this y^2=(4-x^2)d/dx basically says, how does this change if I change x?  You can also differentiate with respect to t instead, or even another variable.  Just depends on the question and the context.