I'm having some issues with the following problem:
Find the slope of the curve $\tan^{-1}{\left(\frac{2x}{y}\right)=\frac{\pi{}x}{y^2}}$ at the point $(1, 2)$.
Since I'm looking for the slope I want to differentiate the function of the curve, and to do that I re-wrote it as: $\tan^{-1}{\left(\frac{2x}{y}\right)=\frac{\pi{}x}{y^2}} \Leftrightarrow {tan}^{-1}{\left(\frac{2x}{y}\right)-\frac{\pi{}x}{y^2}=0}$.
By letting $f(x)=\tan^{-1}{\left(\frac{2x}{y}\right)-\frac{\pi{}x}{y^2}}$, I now can differentiate $f(x)$:
$$\frac{d}{dx}f\left(x\right)=\frac{1}{1+{\left(\frac{2x}{y}\right)}^2}\times{}\frac{d}{dx}\left(\frac{2x}{y}\right)-\frac{d}{dx}\left(\frac{\pi{}x}{y^2}\right)=\frac{1}{1+{\left(\frac{2x}{y}\right)}^2}\times{}\left(\frac{2}{y}\right)-\left(\frac{\pi{}}{y^2}\right)$$
Finally, I can replace $x$ and $y$ with 1 and 2, and after simplifying the expresion I end up with $\frac{1}{2}-\frac{\pi{}}{4}$, but the answer should be $\frac{\pi{}-2}{\pi{}-1}$.
What did I do wrong?
Best Answer
Your work looks okay. Here is what I have.
Note that $$ f(x,y)\equiv 0 \Rightarrow \frac{\partial f}{\partial y} \frac{dy}{dx} + \frac{\partial f}{\partial x}=0 \tag1 $$ Doing the calculus I get $${{2}\over{\left({{4\,x^2}\over{y^2}}+1\right)\,y}}-{{\pi}\over{y^2 }}+\left({{2\,\pi\,x}\over{y^3}}-{{2\,x}\over{\left({{4\,x^2}\over{ y^2}}+1\right)\,y^2}}\right)\frac{dy}{dx}=0\tag2$$
Do the substitutions to get $$\left({{\pi}\over{4}}-{{1}\over{4}}\right)\,\frac{dy}{dx}-{{\pi}\over{4}}+{{1 }\over{2}}=0$$ which gives the answer you want.