Why can't we do substitution in differentiation but is it ok in Taylor series?

Question

I had the same question 10 years ago when I was studying high school. I didn't understand it and I gave up the math. 10 year ago, I needed to work with calculus during work and this question came to find me again. This question is really silly but why?

Why can't I simply substitute $\frac{d(\ln(x))}{dx}$ with $\frac{d(x)}{dx}$?

I.e. let $\ln(x)=x$, as $\frac{dx}{dx}= 1$, why not $\frac{d(\ln(x))}{dx}= 1$?

On the other hand, why can Taylor series such as $\sin(x^2)$ let $x=x^2$ and substitute into Taylor series of $\sin(x)$? In Taylor series, we need to differentiate the function too. Why there is such a difference?

When can I do substitution and why I can't? I can't figure it out.

Answer 1

Just entertaining the notion to show you the mistake in your reasoning rather than justifying the method! If we have this problem:

$$ \frac{d(\ln(x))}{dx} $$ and we let $y=\ln(x) $, which implies that $x=e^y$

then we can rewrite:

$$ \frac{d(\ln(x))}{dx}= \frac{d(y)}{de^y} $$ again, I don't think this is useful to do for any practical purposes, but it highlights the error in your argument.

Answer 2

The issue is with your initial attempt at a substitution. "Let $\ln(x)=x$" makes no sense. Consider, for example, this substitution when the $x$ variable takes the value $x=1$. Then your proposed substitution becomes the contradictory $\ln(1)=1$, or $0=1$.

A substitution should introduce a new variable. Thus "Let $\alpha=\ln(x)$" is fine because when $x=1$, $\alpha$ may take the value $\alpha=0$, and we are not forced to affirm the non-equation $1=0$.

Answer 3

You can substitute $x \rightarrow x^2$ (or rather: use $x^2$ as an argument in the sine function) in Taylor series since it's correct for every real number. However, the formula changes so you can't say that $\sin x$ is identical with $\sin x^2$. $$\sin x = x - \frac{x^3}{3!} + \frac {x^5}{5!} + \ldots,$$ $$\sin x^2 = x^2 - \frac{(x^2)^3}{3!} + \frac {(x^2)^5}{5!} + \ldots$$ If you wanted to obtain the expansion of $\sin x^2$ without using the expansion of $\sin x$ you would have to compute the derivatives of $\sin x^2$.
When you substitute, you need to change every instance of old variable to the new variable (as in the dimebucker91's answer you had to do the change $x \rightarrow e^y$ from the denominator).