# Implicit Differentiation Proof

I'm sure this question has been asked before and yet I can't find it.

Is there a proof of implicit differentiation or is it simply an application of the chain rule? If it's the former, could you give or point me to the proof? If the latter, could you explain exactly how conceptually it works (or point to a link that does so)?

Thank you.

• Does this help: math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/… ? – user21436 Dec 27 '11 at 23:07
• Yes, to some extent, since it says that it's just a special case of the chain rule but it doesn't explain further. Is it because when you are differentiating a mapping with respect to x, you can treat y or any other variables as a function of x and then use the chain rule? That would imply implicit differentiation is generalizable to arbitrarily many other variables. Is that true? – Deets McGeets Dec 27 '11 at 23:13
• Each of the answers below is elucidating. Unfortunately, I can only choose one answer, so I chose the fastest response. – Deets McGeets Dec 27 '11 at 23:43
• Do you want the proof of implicit function theorem? – jdh8 Sep 16 '13 at 16:34
• I too used to have a similar doubt. The way I got over this is, considering an example, $2xy=x^3+y^3$, and remembering that $y$ nothing else than $f(x)$, was to take one side as $h(x)$, in this case $h(x)=2xy$ [does not really matter that it is dependent on $y$]. Then, how would you normally differentiate $h(x) = x^3+y^3$? – Sawarnik Feb 1 '14 at 18:21

It doesn't make any more sense to "prove implicit differentiation" than it does to "prove numbers," but I assume you're asking why implicit differentiation is valid i.e. preserves the truth of equations.

Implicit differentiation is just an application of the chain and other derivative rules to both sides of an equation, with (in the usual case) $y$ an abridgment of $f(x)$. Observe:

$$\frac{d}{dx}g(f(x),x)=\frac{\partial g}{\partial f}(f(x),x)\cdot\frac{df}{dx}(x)+\frac{\partial g}{\partial x} \tag{1}$$

$$\left(g(y,x)\right)'=\frac{\partial g}{\partial y}y'+\frac{\partial g}{\partial x} \tag{2}$$

The first is how you would take the derivative of a function of both $y=f(x)$ and $x$, but notice the difference between $d/dy$ and $\partial/\partial y$, while the latter equation is the same thing in fewer symbols courtesy of implicit differentiation. You always use ID in contexts where you have a function of both dependent and independent variables, and it's valid because of the above analogy - and it indeed generalizes to arbitrarily many dependent (or independent!) variables. For example, if we understand $u$ and $v$ to be functions of $t$ then we can differentiate

$$u^2+tv^2=1\quad\longrightarrow\quad 2uu'+(1\cdot v^2+t\cdot2vv')=0.$$

Implicit differentiation is simply the use of the chain rule to differentiate a function. Often this makes it possible to differentiate a function that is difficult or impossible to separate into the form $y = f(x)$.

For example, consider the function $y = e^{xy}$. There is no way to separate $y$, so it is impossible to take an 'explicit' derivative. However, we can find an 'implicit' derivative in terms of $x$ and $y$ using the chain rule.

How do we do this? Well, simply take the derivative of both sides: $$\frac{d}{dx} y = \frac{d}{dx} e^{xy}$$ The left hand side is simply $dy/dx$, or $y'$. The right hand side looks strange to derive, since the equation depends on both y and x. However, a few applications of the chain rule will clear things right up. First, let the 'inside function' be $u = xy$ and the outside function $e^u$. The derivative is, of course, $$\frac{d}{dx} e^u = \frac{du}{dx} \frac{d}{du} e^u = \frac{du}{dx} e^u$$ Now, using the product rule and then the chain rule, $$\frac{du}{dx} = \frac{d}{dx} xy = (\frac{d}{dx} x)y + x (\frac{d}{dx} y) = (1)y + x(\frac{dy}{dx} \frac{d}{dy}y) = y + x(\frac{dy}{dx}*1) = y + xy'$$ Finally, our implicit derivative is $$y' = (y + xy')e^{xy}$$ and we can easily solve for $y'$.

So you see that implicit differentiation is the result of giving up on separating variables to get a derivative in terms of $x$ only, and cleverly using the chain rule to instead get a derivative in terms of $x$ and $y$.

If you see, for example, $$\cdots + x^3 y^4 + \cdots$$ remember that $y$ is a function of $x$, so it's as if you have $$\cdots + x^3 \big(f(x)\big)^4 + \cdots.$$ Then $$\frac{d}{dx}\left( \cdots + x^3 \big(f(x)\big)^4 + \cdots \right) = \cdots + \left(\frac{d}{dx} x^3\right) \big(f(x)\big)^4 + x^3 \frac{d}{dx} \left(\big(f(x)\big)^4\right) + \cdots$$ (that's the product rule) $$= \cdots + 3x^2 \big(f(x)\big)^4 + x^3 4\big( f(x)\big)^3 \frac{d}{dx} f(x) + \cdots$$ That's the chain rule.

Now write it with $y$ wherever $f(x)$ is (and that means $\dfrac{dy}{dx}$ wherever $\dfrac{d}{dx} f(x)$ is).
The text passage is a discussion about implicit differentiation in calculus, including a question about its proof and explanations of how it works. It does not contain any specific mathematical questions or calculations that require a numerical answer.
no_question
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First, describe the content structure you found in 1-2 sentences.

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# Implicit Differentiation Proof

I'm sure this question has been asked before and yet I can't find it.

Is there a proof of implicit differentiation or is it simply an application of the chain rule? If it's the former, could you give or point me to the proof? If the latter, could you explain exactly how conceptually it works (or point to a link that does so)?

Thank you.

• Does this help: math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/… ? – user21436 Dec 27 '11 at 23:07
• Yes, to some extent, since it says that it's just a special case of the chain rule but it doesn't explain further. Is it because when you are differentiating a mapping with respect to x, you can treat y or any other variables as a function of x and then use the chain rule? That would imply implicit differentiation is generalizable to arbitrarily many other variables. Is that true? – Deets McGeets Dec 27 '11 at 23:13
• Each of the answers below is elucidating. Unfortunately, I can only choose one answer, so I chose the fastest response. – Deets McGeets Dec 27 '11 at 23:43
• Do you want the proof of implicit function theorem? – jdh8 Sep 16 '13 at 16:34
• I too used to have a similar doubt. The way I got over this is, considering an example, $2xy=x^3+y^3$, and remembering that $y$ nothing else than $f(x)$, was to take one side as $h(x)$, in this case $h(x)=2xy$ [does not really matter that it is dependent on $y$]. Then, how would you normally differentiate $h(x) = x^3+y^3$? – Sawarnik Feb 1 '14 at 18:21

It doesn't make any more sense to "prove implicit differentiation" than it does to "prove numbers," but I assume you're asking why implicit differentiation is valid i.e. preserves the truth of equations.

Implicit differentiation is just an application of the chain and other derivative rules to both sides of an equation, with (in the usual case) $y$ an abridgment of $f(x)$. Observe:

$$\frac{d}{dx}g(f(x),x)=\frac{\partial g}{\partial f}(f(x),x)\cdot\frac{df}{dx}(x)+\frac{\partial g}{\partial x} \tag{1}$$

$$\left(g(y,x)\right)'=\frac{\partial g}{\partial y}y'+\frac{\partial g}{\partial x} \tag{2}$$

The first is how you would take the derivative of a function of both $y=f(x)$ and $x$, but notice the difference between $d/dy$ and $\partial/\partial y$, while the latter equation is the same thing in fewer symbols courtesy of implicit differentiation. You always use ID in contexts where you have a function of both dependent and independent variables, and it's valid because of the above analogy - and it indeed generalizes to arbitrarily many dependent (or independent!) variables. For example, if we understand $u$ and $v$ to be functions of $t$ then we can differentiate

$$u^2+tv^2=1\quad\longrightarrow\quad 2uu'+(1\cdot v^2+t\cdot2vv')=0.$$

Implicit differentiation is simply the use of the chain rule to differentiate a function. Often this makes it possible to differentiate a function that is difficult or impossible to separate into the form $y = f(x)$.

For example, consider the function $y = e^{xy}$. There is no way to separate $y$, so it is impossible to take an 'explicit' derivative. However, we can find an 'implicit' derivative in terms of $x$ and $y$ using the chain rule.

How do we do this? Well, simply take the derivative of both sides: $$\frac{d}{dx} y = \frac{d}{dx} e^{xy}$$ The left hand side is simply $dy/dx$, or $y'$. The right hand side looks strange to derive, since the equation depends on both y and x. However, a few applications of the chain rule will clear things right up. First, let the 'inside function' be $u = xy$ and the outside function $e^u$. The derivative is, of course, $$\frac{d}{dx} e^u = \frac{du}{dx} \frac{d}{du} e^u = \frac{du}{dx} e^u$$ Now, using the product rule and then the chain rule, $$\frac{du}{dx} = \frac{d}{dx} xy = (\frac{d}{dx} x)y + x (\frac{d}{dx} y) = (1)y + x(\frac{dy}{dx} \frac{d}{dy}y) = y + x(\frac{dy}{dx}*1) = y + xy'$$ Finally, our implicit derivative is $$y' = (y + xy')e^{xy}$$ and we can easily solve for $y'$.

So you see that implicit differentiation is the result of giving up on separating variables to get a derivative in terms of $x$ only, and cleverly using the chain rule to instead get a derivative in terms of $x$ and $y$.

If you see, for example, $$\cdots + x^3 y^4 + \cdots$$ remember that $y$ is a function of $x$, so it's as if you have $$\cdots + x^3 \big(f(x)\big)^4 + \cdots.$$ Then $$\frac{d}{dx}\left( \cdots + x^3 \big(f(x)\big)^4 + \cdots \right) = \cdots + \left(\frac{d}{dx} x^3\right) \big(f(x)\big)^4 + x^3 \frac{d}{dx} \left(\big(f(x)\big)^4\right) + \cdots$$ (that's the product rule) $$= \cdots + 3x^2 \big(f(x)\big)^4 + x^3 4\big( f(x)\big)^3 \frac{d}{dx} f(x) + \cdots$$ That's the chain rule.

Now write it with $y$ wherever $f(x)$ is (and that means $\dfrac{dy}{dx}$ wherever $\dfrac{d}{dx} f(x)$ is).
user
The text passage is a discussion about implicit differentiation in calculus, including a question about its proof and explanations of how it works. It does not contain any specific mathematical questions or calculations that require a numerical answer.

TYPE: NO_QUESTION