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

1 Answer
Feb 7, 2016

frac{dy}{dx} = frac{sec^2(x-y) - y}{sec^2(x-y) + x}dydx=sec2(xy)ysec2(xy)+x

Explanation:

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

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

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

= xfrac{d}{dx}(y) + yfrac{d}{dx}(x) - sec^2(x-y)frac{d}{dx}(x-y)=xddx(y)+yddx(x)sec2(xy)ddx(xy)

= xfrac{dy}{dx} + y - sec^2(x-y)(1-frac{dy}{dx})=xdydx+ysec2(xy)(1dydx)

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

sec^2(x-y) - y = (sec^2(x-y) + x)frac{dy}{dx}sec2(xy)y=(sec2(xy)+x)dydx

frac{dy}{dx} = frac{sec^2(x-y) - y}{sec^2(x-y) + x}dydx=sec2(xy)ysec2(xy)+x