Implicit Differentiation: Differentiating in Terms of X

[Drakkith]

I'm having some trouble with the terminology used in calculus.

My book states: "Fortunately we don't need to solve an equation for Y in terms of X in order to find the derivative of Y. Instead we can use the method of implicit differentiation. This consists of differentiating both sides of the equation with respect to X and then solving the resulting equation for Y'."

And: "In the examples and exercises of this section it is always assumed that the given equation determines Y implicitly as a differentiable function of X so that the method of implicit differentiation can be applied."

I don't quite understand what they are telling me here. What does "differentiating with respect to X" and "Y as a differentiable function of X" mean?

If it helps explain the above, here's an example equation from the section: X²+Y² = 25

 

[Mark44]

If it helps explain the above, here's an example equation from the section: X²+Y² = 25

Differentiating with respect to x means applying the $\frac d {dx}$ operator. If we differentiate both sides of the equation wrt x, we get:
$\frac d {dx} (x^2 + y^2) = \frac d {dx} 25$
$\Rightarrow \frac d {dx} x^2 + \frac d {dx} y^2 = 0$
$\Rightarrow 2x + \frac d {dx} y^2 = 0$

To differentiate y² with respect to x, we need to use the chain rule, something you asked about in a previous thread.
$\frac d {dx} y^2 = \frac d {dy} y^2 \frac{dy}{dx} = 2y \frac {dy}{dx} = 2y * y'$

 

[Drakkith]

So "differentiating Y² with respect to X" just means we are doing F'(x)?

Thinking more about the second question, I'm guessing that "Y as a differentiable function of X" means that Y = F(x) where F(x) is a differentiable function?

 

[Mark44]

So "differentiating Y² with respect to X" just means we are doing F'(x)?

No. You don't "do" F'(x).
F'(x) is a thing (a noun). Differentiating is an action (a verb).

F'(x) is what you get by differentiating F(x). IOW, $\frac d {dx} F(x) = F'(x)$.

Thinking more about the second question, I'm guessing that "Y as a differentiable function of X" means that Y = F(x) where F(x) is a differentiable function?

Sort of, but you don't need F there. if y is a differentiable function of x, then dy/dx exists.

 

[Drakkith]

Alright, thanks Mark.

 

[NascentOxygen]

I'm having some trouble with the terminology used in calculus.

My book states: "Fortunately we don't need to solve an equation for Y in terms of X in order to find the derivative of Y. Instead we can use the method of implicit differentiation. This consists of differentiating both sides of the equation with respect to X and then solving the resulting equation for Y'. " [...]

If it helps explain the above, here's an example equation from the section: X²+Y² = 25

Taking your sample equation, there are two routes by which you can find $\frac {dy} {dx}$

Method 1: re-arrange the equation to isolate y on one side by itself,

$y\ =\ ±\sqrt{25\ -\ x^2}$

and now differentiate both side with respect to x. (This is the method you most likely would have used.)

Method 2: don't rearrange the equation; keep it as just as it is and differentiate both sides with respect to x. Mark44 has demonstrated Method 2, though hasn't finished it to the point of expressing dy/dx as a function of x only (because often this last step is not necessary). But perhaps as an exercise you could complete it, to show dy/dx as a function of x only. The final answer will be the same as for method 1. https://www.physicsforums.com/threa...rentiating-in-terms-of-x.805803/#post-5058053

The reason your textbook implies that Method 2 is easier/shorter may not yet be apparent. But if your task were to determine the value of $\frac {dy} {dx}$ at a given point, say (3,-4), then Method 2 may turn out easier. For more complicated equations it generally will.

 

[micromass]

If you're still having trouble, then perhaps replace $y$ with $f(x)$ and do what you usually do.

So if you have $x^2 + y^2 = 2$. Then you should see it as $x^2 + f(x)^2 =2$. Then differentiate both sides to get $2x + 2f(x)f^\prime(x) = 0$.

By the way, if you're like me and often forget the quotient rule, then you can do it with implicit differentiation. So if $f(x) = \frac{F(x)}{G(x)}$. Then rewrite it as $f(x)G(x) = F(x)$. Differentiate both sides to get $f^\prime(x)G(x) + f(x)G^\prime(x) = F^\prime(x)$. Isolate $f^\prime(x)$ to get

$$f^\prime(x) = \frac{F^\prime(x) - f(x)G^\prime(x)}{G(x)}= \frac{F^\prime(x) - \frac{F(x)}{G(x)} G^\prime(x)}{G(x)} = \frac{ F^\prime(x)G(x) - F(x)G^\prime(x)}{G(x)^2}.$$

So you can rediscover the formula like this, or just apply it to a special case. Like if you have $f(x) = \frac{x^2 + 1}{2x}$, then just differentiate both sides $2xf(x) = x^2 + 1$.

 

[Drakkith]

Thanks guys. After doing more of these problems it's a little clearer on what's going on.