0
$\begingroup$

Please take a look at my case:

$$ t(x)=1/(1+exp^{-x})\\therefore\ we\ can\ create\ three\ functions: \\f(x)=exp^{-x}\\g(f(x)) = 1+f(x)\\c(g(f(x)) = 1/g(f(x))\\ $$

we know

$$t'(x) = t(x)*(1-t(x))$$

and i think it should be:

$$t'(x) = t(x)*(1-t(x)*exp^{-x})$$

why?

chain rule says, find derivative of each function and multiply together, therefore we should get out this:

$$ t'(x) = -1(1/g(f(x)))^-2*exp^{-x}*exp^{-x}$$

or on different case:

$$ \\t(x) = ((x^2)^2)^2\\ \\f(x) = x^2\\ \\g(f(x)) = f(x)^2\\ \\c(g(f(x)) = g(f(x))^2\\ $$

therefore derivative is:

$$ t'(x) = 8x^7 $$

$\endgroup$
  • $\begingroup$ MathJax hints: if you put backslashes before common functions you get the proper font and spacing, so \exp x gives $\exp x$. When you want multicharacter exponents, put them in braces, so e^{-x} gives $e^{-x}$ instead of e^-x which gives $e^-x$. This works everywhere-thinks in braces are considered a unit. $\endgroup$ – Ross Millikan Mar 29 '18 at 16:20
  • $\begingroup$ @RossMillikan Thanks I fixed it $\endgroup$ – filtertips Mar 29 '18 at 16:22
  • 1
    $\begingroup$ Introducing a different $t(x)$ makes your argument hard to follow. You should start from $t(x)=c(g(f(x)))$ and apply the chain rule to that. It should also be either $\exp(-x)$ or $e^{-x}$, not $exp^{-x}$ $\endgroup$ – Ross Millikan Mar 29 '18 at 16:23
  • $\begingroup$ Related: math.stackexchange.com/questions/78575/… $\endgroup$ – Hans Lundmark Mar 29 '18 at 18:28
0
$\begingroup$

Since $g(y) = 1+y$, we have $g'(y) = 1$ for any $y$, so the middle step of your chain rule is wrong. It should be $t'(x) = -\frac{1}{g(f(x))^2} \cdot 1 \cdot (-e^{-x})$.

$\endgroup$
  • $\begingroup$ Thank you. I have read on wikipedia derivative of e{x} is always e{x}. By the way, where this you get - (minuses) in t'(x). I would appreciate it. $\endgroup$ – filtertips Mar 29 '18 at 16:38

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.