Nash Frank

2022-07-16

The question asks to implicitly differentiate $\mathrm{tan}\left(x+y\right)=x$.
I get the correct answer of $-{\mathrm{sin}}^{2}\left(x+y\right)$. However, the book also says "or you can use $-\frac{{x}^{2}}{{x}^{2}+1}$". I don't see how I can construct a proper right triangle to give/obtain this equivalent answer. Can you assist?

Emilie Reeves

Expert

From this you have: $x+y={\mathrm{tan}}^{-1}x+n\pi \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}1+{y}^{\prime }=\frac{1}{{x}^{2}+1}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime }=....$

Ciara Rose

Expert

Note that from the implicit equation $\mathrm{tan}\left(x+y\right)=x$ you obtain $y=\mathrm{arctan}\left(x\right)-x$ and therefore
$-{\mathrm{sin}}^{2}\left(x+y\right)=-{\mathrm{sin}}^{2}\left(x+\mathrm{arctan}\left(x\right)-x\right)=-{\mathrm{sin}}^{2}\left(\mathrm{arctan}\left(x\right)\right)$
Now by drawing a right triangle with hypotenuse $\sqrt{1+{x}^{2}}$ and opposite side $x$ and adjacent side 1 note that
$\mathrm{sin}\left(\mathrm{arctan}\left(x\right)\right)=\frac{±x}{\sqrt{1+{x}^{2}}}$
and thus
$-{\mathrm{sin}}^{2}\left(\mathrm{arctan}\left(x\right)\right)=-{\left(\frac{±x}{\sqrt{1+{x}^{2}}}\right)}^{2}=\frac{-{x}^{2}}{{x}^{2}+1}$

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