for $x > 0$, find and simplify the derivative of $f(x)=arctan x+arctan(1/x)$. $f(x)=$ (what does your result…

Explanation:

Step1: Recall derivative of arctan(u)

The derivative of $\arctan(u)$ with respect to $x$ is $\frac{u'}{1 + u^{2}}$. For $y_1=\arctan(x)$, $u = x$ and $u'=1$, so $y_1'=\frac{1}{1 + x^{2}}$. For $y_2=\arctan(\frac{1}{x})$, let $u=\frac{1}{x}=x^{-1}$, then $u'=-x^{-2}=-\frac{1}{x^{2}}$.

Step2: Find derivative of $y_2$

Using the chain - rule, $y_2'=\frac{-\frac{1}{x^{2}}}{1+(\frac{1}{x})^{2}}$. Simplify the denominator: $1 + (\frac{1}{x})^{2}=1+\frac{1}{x^{2}}=\frac{x^{2}+1}{x^{2}}$. So $y_2'=\frac{-\frac{1}{x^{2}}}{\frac{x^{2}+1}{x^{2}}}=-\frac{1}{x^{2}+1}$.

Step3: Find derivative of $f(x)$

Since $f(x)=y_1 + y_2$, then $f'(x)=y_1'+y_2'$. Substitute $y_1'$ and $y_2'$: $f'(x)=\frac{1}{1 + x^{2}}-\frac{1}{1 + x^{2}}$.

Answer:

$0$

The result $f'(x) = 0$ for $x>0$ tells us that the function $f(x)=\arctan(x)+\arctan(\frac{1}{x})$ is a constant function for $x > 0$.