The proof of $\frac{d}{dx}(\ln{x}) = \frac{d}{dx}(\ln{ax})$ is pretty simple: use chain rule, then the a constants cancel, leaving you with 1/x, which is also the derivative of ln(x). -- Logically, how does this make sense? The a constant would have an effect on the graph of ln(x). And if it does not, or if the constant only serves to vertically translate ln(x), why is that? Thank you for your help.
121471
The integral of $\frac{1}{2x}$ is $\frac{\ln(x)}{2}$, but can't it also be $\frac{\ln(2x)}{2}$ or $\frac{\ln(3x)}{2}$?  Is there a special reason for $\ln(Ax)$ to have identical derivatives?
34545
I have tried to let$f(a)=\int_0^1\frac{\ln(1-ax)}{x} dx$, which means     $f'(x)=-a\int_0^1 \frac{1}{x(1-ax)} dx=a\int_0^1 \left[\frac{a}{ax-1}-\frac{1}{x}\right] dx$.   However, I am stuck at here.
7827
I wanted to verify that for definition of the derivative it is true that:  $$ \lim_{x \to a}\frac{f(x)-f(a)}{x-a}= \lim_{h\to 0}\frac{f(a+h) -f(a)}{h}$$  If I denote $h=x-a$,  we can  let $x\to a$, this means that $h \to 0$. Also notice that $f(x)=f(a+h)$ $$ \lim_{x \to a}\frac{f(x)-f(a)}{x-a}=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$$ As desired. Is the main idea here really just to use a variable substitution? I suppose there is some formal theorem I also use here that when $f(x) \to y$  we can let $g(f(x)) \to g(y)$ if the function $g$ is continuous.
49871
When we take a derivative of a function where the power rule applies, e.g. $x^3$, we multiply the function by the exponent and subtract the current exponent by one, receiving $3x^2$. Using this method, why is it that the derivative for $e^x$ equal to itself as opposed to $xe^{x-1}$? I understand how the actual derivative is derived (through natural logs), but why can't we use the original method to differentiate? Furthermore, why does the power rule even work? Thank you all in advance for all of your help.
57478
What would be the $n$th derivative of   $ f (x) = x ^ x$  I have reached the fifth derivative, very long indeed but I see no pattern that will help me find a general expression.  \begin{align*} \frac{df}{dx} &amp;= x^x(1+\ln(x))\\ \frac{d^2f}{dx^2} &amp;= x^x\left(\frac{1}{x}+1+2\ln(x)+\ln(x)^2\right)\\ \frac{d^3f}{dx^3} &amp;= x^x\left( \frac{-1}{x^2}+\frac{3}{x} + \ln(x)^3 + 3 \ln(x)^2 + \frac{3\ln(x)}{x} + 3\ln(x) + 1 \right)\\ \frac{d^4f}{dx^4} &amp;= x^x\left( \frac{2}{x^3}-\frac{1}{x^2}-\frac{4\ln(x)}{x^2}+\frac{6}{x}+\ln(x)^4+4\ln(x)^3\right.\\ &amp;\qquad\qquad\left.+\frac{6\ln(x)^2}{x}+6\ln(x)^2+\frac{12\ln(x)}{x}+4\ln(x)+1 \right)\\ \frac{d^5f}{dx^5} &amp;= x^x\left( \frac{-6}{x^4}+\frac{10\ln(x)}{x^3}+\frac{5}{x^2}-\frac{10\ln(x)^2}{x^2}+\frac{10}{x}+\ln(x)^5+5\ln(x)^4\right.\\ &amp;\qquad\qquad\left.+\frac{10\ln(x)^3}{x}+10\ln(x)^3+\frac{30\ln(x)^2}{x}+10\ln(x)^2+\frac{30\ln(x)}{x}+5\ln(x)+1 \right) \end{align*}
95777
I know that $\int \frac{1}{x}~dx = \ln|x| + C$ and I know the antiderivative method works for all powers of $x$ except $-1$. But why is that the case? I am still in high school and teachers aren't really helpful with these questions.   Edit: I do realize dividing by zero is meaningless. That's not the question. Just because dividing by zero is meaningless doesn't mean mathematicians just choose to ignore it and work just make up a new answer. There has to be an explanation, and that's what I am asking for. The explanation.
926
How do I show that $\ln(x) = \lim_{n\to\infty} n (x^{1/n} - 1)$?  I ran into this identity on . I haven't been able to find any proof online and my efforts to get from $\ln(x) := \int_1^x \frac{\mathrm dt}t$ to that limit have been a failure.
107437
I'm trying to write a proof for this, but I don't know how to begin. Any help appreciated.  Suppose $g$ is a function  such that $g'(x) = \frac{1}{x}$ for all $x &gt; 0$ and $g(1) = 0$. Prove that $g(xy) = g(x) + g(y)$ for all $x, y &gt; 0$.    Let $h(x) = g(x,y)$
73841
So taking the derivative of the latter half we get  $$\left(\frac{2}{x^{3}}\right)e^{\frac{-1}{x^{2}}}$$ by chain rule. Im not sure how to find the derivative at x=0, and show that it equals 0 as it only produces more '1/0' expressions. Additionally I assume to show it is not analytic at $x_0=0$, we use the taylor series for $e^{x}$ which is $$1+x+\frac{x^2}{2}+\cdots+\frac{x^{n}}{n!}$$ Where we substitute $\frac{-1}{x^{2}}$ for x to get  $$e^{\frac{-1}{x^{2}}}=1-\frac{1}{x^{2}}+\frac{1}{x^{4}2!}-\frac{1}{x^{6}3!}\cdots$$ So when we plug in x=0 this is invalid and thus is not analytic.  Is my latter attempt sufficient or do I need to add more? And how do I reach the conclusion that the function is differentiable at x=0.
27738
Prove that      $$ \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx = \ln\left(\frac{a}{b}\right) $$   My Attempt:  Define the function $I(a,b)$ as  $$ I(a,b) = \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx $$  Differentiate both side with respect to $a$ to get  $$ \begin{align} \frac{dI(a,b)}{da} &amp;= \int_{0}^{\infty}\frac{0-e^{-ax}(-x)}{x}\,dx\\ &amp;= \int_{0}^{\infty}e^{-ax}\,dx\\ &amp;= -\frac{1}{a}(0-1)\\ &amp;= \frac{1}{a} \end{align} $$  How can I complete the proof from here?
99757
To Prove : $$ \frac{f'(c)}{f(c)} = \frac{1}{a-c} + \frac{1}{b-c} $$ I think I should proceed in following way :  Define $g(x) = \ln|f(x)| + \ln|x-a| + \ln|b-x|$ such that $$g'(x) = \frac{f'(x)}{f(x)} - \frac{1}{a-x} - \frac{1}{b-x} $$ Then, using Rolle's Theorem , I have to prove that : $\exists c \in (a,b)$ such that $g'(c) = 0$ But I am not able to solve it any further .
112183
Let $f =f(x(t),y(t))$  Then the total derivative of f with respect to t is given by:  $$\frac {}{} = \frac {}{} + \frac {}{}$$  Say still we have  $f=f(x(t),y(t))$. What is then $\frac {\partial^2 f}{ \partial t^2} $? Is it just $$ \frac {\partial f}{ \partial x} \frac {d^2x}{dt^2}+ \frac {\partial f }{\partial y} \frac {d^2y}{dt^2} $$ ? (Just by applying chain rule again?)  Also, this seems to be related to the Jacobian  J=J(f). Is this correct? If so, can we define $\frac{\partial^2 f}{ \partial t^2}$ as $JJ^T$ , where $J^T$ is the transpose of $J$?
105572
Let's assume the functions in question are continuously differentiable. If $f(g(x)) = f(x)g(x)$, then: $$f'(g(x))g'(x) = f'(x)g(x)+f(x)g'(x)$$ I attempted to find a particular solution by letting $g(x) = x$. Then, $$f'(x) = xf'(x) + f(x)$$ $$df = xdf + fdx$$ $$\frac{df}{f} = \frac{dx}{1-x}$$ $$\ln(f(x)) = -\ln(1-x)+C$$ $$f(x) = \frac {A}{1-x}$$ Where $A$ is a constant. This solution was fairly unsatisfying to me, and upon checking it, I see it doesn't actually work, as letting $g(x) = x$ leads to $f(x) = xf(x)$. I was wondering if there were better answers, or at least a less trivial way to go about the problem.
61261
This is a very common proof, but there is one point, which many just seem to take for granted.  At one point I'll use the mean value theorem to say  $$ w.l.o.g.\quad  x &gt; y \quad x,y \in (a,b) $$   And from the mean value theorem I get a term like this  $$\frac{f(x)-f(y)}{x-y} = f'(\xi) \leq L $$  But people often conclude $$|f(x)-f(y)| \leq L|x-y|$$  Is that step always true? i.e. is it always true that   $$ x = y \rightarrow |x|=|y| $$
84672
Letting $x=\tan(u)$, $$\int\frac{1}{1+x^2} \, dx=\int\frac{1}{1+\tan(u)^2}\sec^2(u) \, du=u+k=\arctan(x)+k$$ Also,  $$\int\frac{1}{1+x^2} \, dx=\int\frac{1}{(x-i)(x+i)} \, dx=\frac{1}{2i}\int\frac{1}{(x-i)}-\frac{1}{(x+i)} \, dx$$ $$=\frac{1}{2i}\left(\ln(x-i)-\ln(x+i)\right)+c$$ $$=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+c$$ Giving $$\arctan(x)=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+q$$  Why is this correct? What is the nature of $q$ (is it 'flexible' so the equality doesn't really mean much)?  I think it probably has something to do with the relationship between $\log(z)$ and $\arg(z)$, but $\arg(z\pm i)$ is hard to calculate neatly.
72842
How do I derive:   $\ln(-1)=i\pi$ and   $\ln(-x)=\ln(x)+i\pi$  for $x&gt;0$ and  $x \in\mathbb R$  Thanks for any and all help!
34337
Compute the taylor series of $\ln(1+x)$  I've first computed derivatives (up to  the 4th) of ln(1+x)  $f^{'}(x)$ = $\frac{1}{1+x}$  $f^{''}(x) = \frac{-1}{(1+x)^2}$ $f^{'''}(x) = \frac{2}{(1+x)^3}$ $f^{''''}(x) = \frac{-6}{(1+x)^4}$ Therefore the series: $\ln(1+x) = f(a) + \frac{1}{1+a}\frac{x-a}{1!} - \frac{1}{(1+a)^2}\frac{(x-a)^2}{2!} +  \frac{2}{(1+a)^3}\frac{(x-a)^3}{3!} -  \frac{6}{(1+a)^4}\frac{(x-a)^4}{4!} + ...$  But this doesn't seem to be correct. Can anyone please explain why this doesn't work?  The supposed correct answer is:  As $\ln(1+x) = \int (\frac{1}{1+x})dx$ $\ln(1+x) = \Sigma_{k=0}^{\infty} \int (-x)^k dx$