dy/dx
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(x^m)^((x^m)^(x^m))=(y^n)^((y^n)^(y^n))
(y^n)^((y^n)^(y^n))-(x^m)^((x^m)^(x^m))=0
(y^n)^((y^n)^(y^n))-(x^m)^((x^m)^(x^m))=0
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(x^m)^((x^m)^(x^m))=(y^n)^((y^n)^(y^n))
(y^n)^((y^n)^(y^n))-(x^m)^((x^m)^(x^m))=0
Cool. I can't wait for Calculus. Excited.
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Frankly, I don't understand MathLover1's answer! I do understand that MathLover1 changed the problem slightly. (x^m)^(x^m)^(x^m) is ambiguous. It could be either [(x^m)^(x^m)]^(x^m), which is what MathLover1 changed it to, or it could be (x^m)^[(x^m)^(x^m)], which is much more difficult.
For example, 2^3^2 is ambiguous. It could be (2^3)^2= 8^2= 64[ or it could be 2^{3^2}= 2^9= 1024.
(x^a)^b= x^{ab} so (x^m)^(x^m)= x^{mx^m} and
[(x^m)^(x^m)]^(x^m)= (x^{mx^m})^{x^m}= x^{m(x^m)(x^m)}= x^{mx^{2m}}.
To differentiate that, let \(u= mx^{2m}\) and we have \(x^u\).
Since u is a function of x, we need to use "logarithmic differentiation". Let \(y= x^u\). Then \(ln(y)= ln(x^u)= u ln(x)\). The derivative of \(ln(y)\) with respect to x is \(\frac{1}{y}\frac{dy}{dx}\).
On the right, use the product rule. \(\frac{duln(x)}{dx}= \frac{du}{dx}ln(x)+ u\frac{dln(x)}{dx}\).
Since \(u= mx^{2m}\), \(\frac{du}{dx}= 2m^2x^{2m- 1}\). Of course \(\frac{dln(x)}{dx}= \frac{1}{x}\). So \(\frac{duln(x)}{dx}= 2m^2x^{2m- 1}ln(x)+ mx^{2m}/x= x^{2m-1}(2m^2ln(x)+ m)\).
So we have \(\frac{1}{y}\frac{dy}{dx}= x^{2m-1}(2m^2ln(x)+ m\).
Since \(y= x^u= x^{mx^{2m}}\)
\(\fra{dy}{dx}= x^{mx^{2m}+ 2m- 1}(2m^2ln(x)+ m))\).
For example, 2^3^2 is ambiguous. It could be (2^3)^2= 8^2= 64[ or it could be 2^{3^2}= 2^9= 1024.
(x^a)^b= x^{ab} so (x^m)^(x^m)= x^{mx^m} and
[(x^m)^(x^m)]^(x^m)= (x^{mx^m})^{x^m}= x^{m(x^m)(x^m)}= x^{mx^{2m}}.
To differentiate that, let \(u= mx^{2m}\) and we have \(x^u\).
Since u is a function of x, we need to use "logarithmic differentiation". Let \(y= x^u\). Then \(ln(y)= ln(x^u)= u ln(x)\). The derivative of \(ln(y)\) with respect to x is \(\frac{1}{y}\frac{dy}{dx}\).
On the right, use the product rule. \(\frac{duln(x)}{dx}= \frac{du}{dx}ln(x)+ u\frac{dln(x)}{dx}\).
Since \(u= mx^{2m}\), \(\frac{du}{dx}= 2m^2x^{2m- 1}\). Of course \(\frac{dln(x)}{dx}= \frac{1}{x}\). So \(\frac{duln(x)}{dx}= 2m^2x^{2m- 1}ln(x)+ mx^{2m}/x= x^{2m-1}(2m^2ln(x)+ m)\).
So we have \(\frac{1}{y}\frac{dy}{dx}= x^{2m-1}(2m^2ln(x)+ m\).
Since \(y= x^u= x^{mx^{2m}}\)
\(\fra{dy}{dx}= x^{mx^{2m}+ 2m- 1}(2m^2ln(x)+ m))\).
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Your LaTex is not appearing as LaTex. Take a look at your replies involving this style of typing.
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I wasn't using Latex because the original post did not use Latex.
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I cannot read what you typed. Take a look at your reply.
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I have tried \(x^2\), [math]x^2[/math], and $x^2$. None seem to work. What should I do?
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just use wolframalpha, type equation or expression, copy and paste image of input
here is a link:
https://www.wolframalpha.com/input/?i=(1+++sec(x))/(tan(x)+++sin(x)+)=+csc(x)
here is a link:
https://www.wolframalpha.com/input/?i=(1+++sec(x))/(tan(x)+++sin(x)+)=+csc(x)
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Wolfram does not provide a screen for typing math solutions to convert to LaTex form. The tiny space bar allows for typing certain questions and, of course, most equations.
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that is true, but to convert to LaTex form is not a point
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Do you know a good app or site that converts to LaTex?
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https://mathpix.com/
Snip can convert images into LaTeX for inline equations, block mode equations, and numbered equations. Snip also supports some text mode LaTeX, like the tabular environment.
scroll down and watch videos
but it's not free for everyone, only for students
Up to 50 Snips
Up to 20 PDF pages
Sign up with your school email (.edu) and get more free Snips and PDFs!
here are 3 Free Word To LaTeX Converter Software For Windows
https://www.ilovefreesoftware.com/15/featured/free-word-to-latex-converter-software-windows.html
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I will visit the links provided.
