I have the following question & its answer but I do not understand how some parts were obtained - Q&A, Previous Q
Where does the $1+dy/dx (x-1)$ come from?
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Please study our guide to new askers, and improve the question. – Jyrki Lahtonen Aug 13 '18 at 05:35
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Hint: Assuming $y=y(x)$ then we get by the chain and product rule
$4x+2y(x)y'(x)+2y(x)+2xy'(x)=0$ Can you finish?
Dr. Sonnhard Graubner
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I've already worked that out to be $-2x+y/x+y$ that's where I apply the bottom bit and get $-(3/2)$ I don't get where the $(x-1)$ comes from. – Daniel Jusere Aug 12 '18 at 10:00
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ok so you don't have any problem with calculating $\frac{dy}{dx}$
you want to find the tangent at the curve at the point (1,1)
the equation of a line passing through the point (1,1) is given by
$$(y-1)=m(x-1)$$ where m is the slope of the line
now to calculate$$ m=slope=\frac{dy}{dx}\big|_{(1,1)}$$ $$ m=slope=-\frac{2x+y}{x+y}\big|_{(1,1)}$$ $$ m=slope=-\frac{2(1)+(1)}{(1)+(1)}=\frac{-3}{2}$$
thus equation of tangent $$(y-1)=-\frac{3}{2}(x-1)$$ $$y-1=-\frac{3}{2}x+\frac{3}{2}$$ $$y=-\frac{3}{2}x+\frac{5}{2}$$
Deepesh Meena
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That makes perfect sense, that’s exactly what I was looking for, thank you. – Daniel Jusere Aug 12 '18 at 10:49