Suppose I have differentiable function $g(x+y)$ and $h(y)$ and define $z = x+y$. Also, I assume that for any $x_1,y_1$ there exists $x_2,y_2$ such that $g(x_1+y_1) = g(x_2+y_2)$. Then, I want to take derivate of $$f(z,y) = g(x+y)h(y)$$ with respect to $y$ and obtain the result like $$g(x+y)h'(y).$$ I want to know if it is possible. My argument is that whenever I change $y$, I can always find $x'$ to offset the effect of the change in $y$ on function $g$.
Asked
Active
Viewed 67 times
0
-
I pressume that $x_1\neq x_2$, and $y_1\neq y_2$? or perhaps $z_1\neq z_2$? Notice that they are not equivalent. – Weierstraß Ramirez Oct 12 '20 at 08:23
1 Answers
0
This seems to not be true in general, perhaps you are looking for a missing assumption?
For instance consider: $g(z)=z^2=(x+y)^2$. Notice that we can then take $x_1=2$, $y_1=-1$, $x_2=-2$, $y_2=1$. Hence $g(x_1+y_1)=g(x_2+y_2)$. The derivative that you are looking for is then:
$$\partial f(x_1,y_1) / \partial y=2(x_1+y_1)h(y)+(x_1+y_1)^2h'(y)=2h(y)+h'(y)$$
Which is in general, not equal to $h'(y)$.
Weierstraß Ramirez
- 1,014
-
I know it won't be generally true. To be exact, I am asking if the following can be true: Let $y_n \rightarrow y$ and $x_n \rightarrow x$ such that $g(x_n,y_n) = g(x_k,y_k)$ for all $n\ne k$. Then, I define derivative as $$\lim_n \frac{f(x_n,y_n)-f(x,y)}{y_n - y} = \lim_n \frac{g(x_0,y_0){h(y_n)-h(y)}}{y_n-y} = g(x_0,y_0)h'(y). $$ – user1292919 Oct 12 '20 at 08:55
-
Can I think of $x_n$, $y_n$ as sequences? $x\to x_0$ when $n\to \infty$, for instance? And $n\in \mathbb{N}$? – Weierstraß Ramirez Oct 12 '20 at 09:22
-
You can think of $x_n$ and $y_n$ as sequences $n$ natural number including zero. But I used $x_0$ and $y_0$ as the first element of the sequences. – user1292919 Oct 12 '20 at 11:41