Find the derivative of \dfrac{\sin^2 x}{1+\cot x}+\dfrac{\cos^2 x}{1+\tan x}.

anemone Paris la ville de l'amour Staff member Administrator Moderator Math Helper Oct 26, 2018 #1 Jan 28, 2018 180 218 43 Oct 26, 2018 #1 Find the derivative of \(\displaystyle \dfrac{\sin^2 x}{1+\cot x}+\dfrac{\cos^2 x}{1+\tan x}\). Likes: MarkFL

MarkFL La Villa Strangiato Staff member Administrator Moderator Math Helper Oct 26, 2018 #2 Jan 25, 2018 3,463 4,254 113 St. Augustine Oct 26, 2018 #2 I think I would begin by looking at ways to simplify first, before differentiating: \(\displaystyle f(x)=\frac{\sin^2(x)}{1+\cot(x)}+\frac{\cos^2(x)}{1+\tan(x)}=\frac{\sin^3(x)}{\sin(x)+\cos(x)}+\frac{\cos^3(x)}{\cos(x)+\sin(x)}=\frac{\sin^3(x)+\cos^3(x)}{\sin(x)+\cos(x)}=\sin^2(x)+\sin(x)\cos(x)+\cos^2(x)=1+\frac{1}{2}\sin(2x)\) And so: \(\displaystyle f'(x)=\cos(2x)\) Likes: anemone

I think I would begin by looking at ways to simplify first, before differentiating: \(\displaystyle f(x)=\frac{\sin^2(x)}{1+\cot(x)}+\frac{\cos^2(x)}{1+\tan(x)}=\frac{\sin^3(x)}{\sin(x)+\cos(x)}+\frac{\cos^3(x)}{\cos(x)+\sin(x)}=\frac{\sin^3(x)+\cos^3(x)}{\sin(x)+\cos(x)}=\sin^2(x)+\sin(x)\cos(x)+\cos^2(x)=1+\frac{1}{2}\sin(2x)\) And so: \(\displaystyle f'(x)=\cos(2x)\)

anemone Paris la ville de l'amour Staff member Administrator Moderator Math Helper Oct 27, 2018 #3 Jan 28, 2018 180 218 43 Oct 27, 2018 #3 Very well done, Mark! This problem is hard to differentiate if one opts to start differentiating and then simplified later...but if we simplify it first, then what we need to differentiate is only the sine function. Thanks for participating, Mark! Likes: MarkFL

Very well done, Mark! This problem is hard to differentiate if one opts to start differentiating and then simplified later...but if we simplify it first, then what we need to differentiate is only the sine function. Thanks for participating, Mark!