How do you implicitly differentiate #-1=xy-tan(x-y) #?

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Answer 1 · 2016-02-07T02:29:05

#frac{dy}{dx} = frac{sec^2(x-y) - y}{sec^2(x-y) + x}#

#-1 = xy - tan(x-y)#

#frac{d}{dx}(-1) = frac{d}{dx}(xy - tan(x-y))#

#0 = frac{d}{dx}(xy) - frac{d}{dx}(tan(x-y))#

# = xfrac{d}{dx}(y) + yfrac{d}{dx}(x) - sec^2(x-y)frac{d}{dx}(x-y)#

# = xfrac{dy}{dx} + y - sec^2(x-y)(1-frac{dy}{dx})#

Now we just have to "shift terms" to make #frac{dy}{dx}# the subject of the formula.

#sec^2(x-y) - y = (sec^2(x-y) + x)frac{dy}{dx}#

#frac{dy}{dx} = frac{sec^2(x-y) - y}{sec^2(x-y) + x}#