# Chain rule in multivariable calculus

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.

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.

• 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? – cgnx Apr 3 '14 at 12:32
• @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. – Mark Fantini Apr 3 '14 at 12:50
• 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? $(\frac{\partial f}{\partial x})'=(\frac{\partial f}{\partial x})'+(\frac {\partial f}{\partial y} \cdot\varphi'(x))'=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial x \partial y} \varphi '(x)+\frac{\partial^2 f}{\partial y \partial x}\varphi'(x)+\frac{\partial^2 f}{\partial y^2}\varphi'(x)+\frac{\partial f}{\partial y}\varphi''(x)$ – cgnx Apr 3 '14 at 13:06
• 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. – Mark Fantini Apr 3 '14 at 13:19
• good! thank you for very detailed explanation, I appreciate the fact that you showed it me generally so I understood the concept. – cgnx Apr 3 '14 at 13:22

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.