How do you implicitly differentiate #-3=cos(y-x)/x#?
1 Answer
Mar 16, 2016
Explanation:
Differentiate both sides w.r.t.
#frac{"d"}{"d"x}(3) = frac{"d"}{"d"x}( cos(y-x)/x )#
#0 = frac{ xfrac{"d"}{"d"x}( cos(y-x) ) - cos(y-x)frac{"d"}{"d"x}(x) }{x^2}#
#0 = x(-sin(y-x))frac{"d"}{"d"x}(y-x) - cos(y-x)#
#0 = xsin(y-x)(frac{"d"y}{"d"x}-1) + cos(y-x)#
Now, we just have to make
#xsin(y-x) - cos(y-x) = xsin(y-x)frac{"d"y}{"d"x}#
#frac{"d"y}{"d"x} = frac{xsin(y-x) - cos(y-x)}{xsin(y-x)}#
#= 1 - cot(y-x)/x#