1
$\begingroup$

I am studying notes from seminar and I don't quite understand some steps. We were evaluating implicit function $f(x,y)=e^{xy}+\sin y +y^2 =1 $ at point $[2,0]$. After checking the conditions we reached the conclusion that such function $y= \varphi(x)$ exists. And now comes the problem, I don't know how the derivative of such function is taken. I know there is chain rule $$\frac{\partial h}{\partial x_k}(\vec{a})=\sum_{j=1}^{p}\frac{\partial f}{\partial y_j}(\vec{b})\frac{\partial g_j}{\partial x_k}(\vec{a}).$$ But I am still struggling with the application of this rule. In this case that for $f(x,\varphi(x))=1$ we would get $$\frac{\partial f}{\partial x}\cdot1+\frac {\partial f}{\partial y}\cdot \varphi'(x)=0.$$ How did the number 1 appear there and same with $\varphi'(x)$? The second derivative is even more overwhelming for me.

$\endgroup$
2
$\begingroup$

Consider the general case. You have $f: \mathbb{R}^2 \to \mathbb{R}$ with $f(h(x),g(x))$. This can be viewed as a one-variable function $y: \mathbb{R} \to \mathbb{R}$, therefore $y(x) = f(h(x),g(x))$. Differentiating: $$\frac{dy}{dx} = \frac{\partial f}{\partial x_1} \frac{dh}{dx} + \frac{\partial f}{\partial x_2} \frac{dg}{dx} = \frac{\partial f}{\partial x} h'(x) + \frac{\partial f}{\partial y} g'(x),$$ where I've briefly denoted the first coordinate by $x_1$ and the second by $x_2$. When you take $h(x) = x$ you get $h'(x) =1$, obtaining that expression.

$\endgroup$
  • $\begingroup$ thank you, I am trying to process it and apply it for the second derivative, just a quick question, what happened to the right side when the function is in my case $f(x,\varphi(x))=1$? I just take derivative of the right side first with respect to x and then y, so I get 0+0=0? $\endgroup$ – cgnx Apr 3 '14 at 12:32
  • $\begingroup$ @D.N I'm sorry, I don't think I understand but I'll try. Still using the general expression, what the equation $f(x, \varphi(x)) = 1$ is saying is that $y(x) \equiv 1$, therefore $f(h(x),g(x)) \equiv 1$. When you differentiate both sides you get the above for the left side and zero for the right side, because differentiating a constant yields zero. $\endgroup$ – Mark Fantini Apr 3 '14 at 12:50
  • $\begingroup$ I am sorry my question is confusing, but basically you answered what I wanted to know. I think I am starting to understand, now can you check if I got it right for the second derivative? $(\frac{\partial f}{\partial x})'=(\frac{\partial f}{\partial x})'+(\frac {\partial f}{\partial y} \cdot\varphi'(x))'=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial x \partial y} \varphi '(x)+\frac{\partial^2 f}{\partial y \partial x}\varphi'(x)+\frac{\partial^2 f}{\partial y^2}\varphi'(x)+\frac{\partial f}{\partial y}\varphi''(x)$ $\endgroup$ – cgnx Apr 3 '14 at 13:06
  • $\begingroup$ Almost. For the term $$\frac{\partial^2 f}{\partial y^2}$$ you need to multiply by $\varphi'(x)$ again. It is $$\frac{\partial^2 f}{\partial y^2} (\varphi' (x))^2.$$ Although we are assuming these functions are $C^2$, notice that $$\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right), $$ that is, you differentiate first in $x$ and second in $y$. Consider accepting this answer if it helped you. =) Best wishes. $\endgroup$ – Mark Fantini Apr 3 '14 at 13:19
  • 1
    $\begingroup$ good! thank you for very detailed explanation, I appreciate the fact that you showed it me generally so I understood the concept. $\endgroup$ – cgnx Apr 3 '14 at 13:22
1
$\begingroup$

If I rpoperly understand your question, you have $$f(x,y)=e^{xy}+\sin y +y^2 =1$$ This obviously define an implict relation between $x$ and $y$; so, $y=\varphi(x)$.

Now, comes the problem of the derivative. What you must write is the total derivative of $f(x,y)$ (which is zero). For this, you need to compute $f'_x$ and $f'_y$ and then $y'_x=-\frac {f'_x} {f'_y}$.

In your case, $f'_x=y e^{x y}$, $f'_y=x e^{x y}+2 y+\cos (y)$. At $[2,0]$ which is along the curve, you then have $f'_x=1$ and $f'_y=3$; so, at this point, $y'_x=-\frac {1} {3}$.

Is this clarifying things to you ? If not, just post.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.