You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
prove with rules of differntiation
- Thread starter dbob85
- Start date
You have the first part OK.
\(\displaystyle \L\\\frac{d}{dx}[tan^{-1}(x)]=\frac{1}{1+x^{2}}\)
Now, the second part, notice the chain in the expression.
\(\displaystyle \L\\\frac{d}{dx}[\frac{1}{x}]=\frac{-1}{x^{2}}\)
Sub 1/x in for x in the arctan differential, getting:
\(\displaystyle \L\\\frac{1}{1+\frac{1}{x^{2}}}\)
Chain rule:
\(\displaystyle \L\\\frac{-1}{x^{2}}\cdot\frac{1}{1+\frac{1}{x^{2}}}\)
Do the algebra, this reduces to:
\(\displaystyle \L\\\frac{-1}{1+x^{2}}\)
So, you have:
\(\displaystyle \L\\\frac{1}{1+x^{2}}-\frac{1}{1+x^{2}}=0\)
\(\displaystyle \L\\\frac{d}{dx}[tan^{-1}(x)]=\frac{1}{1+x^{2}}\)
Now, the second part, notice the chain in the expression.
\(\displaystyle \L\\\frac{d}{dx}[\frac{1}{x}]=\frac{-1}{x^{2}}\)
Sub 1/x in for x in the arctan differential, getting:
\(\displaystyle \L\\\frac{1}{1+\frac{1}{x^{2}}}\)
Chain rule:
\(\displaystyle \L\\\frac{-1}{x^{2}}\cdot\frac{1}{1+\frac{1}{x^{2}}}\)
Do the algebra, this reduces to:
\(\displaystyle \L\\\frac{-1}{1+x^{2}}\)
So, you have:
\(\displaystyle \L\\\frac{1}{1+x^{2}}-\frac{1}{1+x^{2}}=0\)
Hello, dbob85!
Galactus is correct . . . Let me use a slightly different approach.
We have: \(\displaystyle \L\,f(x) \:=\:\arctan(x) \,+ \,\arctan\left(x^{-1}\right)\)
Then: \(\displaystyle \L\:\frac{dy}{dx} \;= \;\frac{1}{1\,+\,x^2} \,+\,\frac{1}{1\,+\,(x^{-1})^2}\,\cdot\left(-x^{-2}\right)\)
. . . . \(\displaystyle \L\:\frac{dy}{dx}\;= \;\frac{1}{1\,+\,x^2}\,-\,\frac{x^{-2}}{1\,+\,x^{-2}}\)
Multiply the second fracton by \(\displaystyle \L\frac{x^{^2}}{x^{^2}}:\)
. . \(\displaystyle \L\:\frac{dy}{dx}\;=\;\frac{1}{1\,+\,x^2} \,-\,\frac{x^{-2}}{1\,+\,x^{-2}}\cdot\frac{x^2}{x^2} \;=\;\frac{1}{1\,+\,x^2} \,-\,\frac{1}{x^2\,+\,1}\;=\;0\)
Galactus is correct . . . Let me use a slightly different approach.
Prove that: \(\displaystyle \L\,\frac{d}{dx}\left[\tan^{-1}(x)\,+\,\tan^{-1}\left(\frac{1}{x}\right)\right] \:= \:0\)
We have: \(\displaystyle \L\,f(x) \:=\:\arctan(x) \,+ \,\arctan\left(x^{-1}\right)\)
Then: \(\displaystyle \L\:\frac{dy}{dx} \;= \;\frac{1}{1\,+\,x^2} \,+\,\frac{1}{1\,+\,(x^{-1})^2}\,\cdot\left(-x^{-2}\right)\)
. . . . \(\displaystyle \L\:\frac{dy}{dx}\;= \;\frac{1}{1\,+\,x^2}\,-\,\frac{x^{-2}}{1\,+\,x^{-2}}\)
Multiply the second fracton by \(\displaystyle \L\frac{x^{^2}}{x^{^2}}:\)
. . \(\displaystyle \L\:\frac{dy}{dx}\;=\;\frac{1}{1\,+\,x^2} \,-\,\frac{x^{-2}}{1\,+\,x^{-2}}\cdot\frac{x^2}{x^2} \;=\;\frac{1}{1\,+\,x^2} \,-\,\frac{1}{x^2\,+\,1}\;=\;0\)