Chain rule in multivariable calculus

Question

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.

Answer 1

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.

Answer 2

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.