What is the implicit derivative of #1=xy-cosy#?

Question

Christopher Allers · Accepted Answer

#frac{"d"y}{"d"x} = -frac{y}{x + sin(y)}#

Implicit differentiation is basically an application of the chain rule.

#frac{"d"}{"d"x}(f(y)) = frac{"d"}{"d"y}(f(y)) frac{"d"y}{"d"x}#

#= f'(y) frac{"d"y}{"d"x}#

So in this problem, we take the derivative of #x# on both sides.

#frac{"d"}{"d"x}(1) = frac{"d"}{"d"x}(xy - cos(y))#

#0 = frac{"d"}{"d"x}(xy) - frac{"d"}{"d"x}(cos(y))#

#= [yfrac{"d"}{"d"x}(x)+xfrac{"d"}{"d"x}(y)] - frac{"d"}{"d"y}(cos(y))frac{"d"y}{"d"x}#

#= y + x frac{"d"y}{"d"x} - (-sin(y))frac{"d"y}{"d"x}#

Now, we just have to make #frac{"d"y}{"d"x}# the subject of formula. Begin by bringing all the terms containing #frac{"d"y}{"d"x}# to one side.

#-y = x frac{"d"y}{"d"x} + sin(y)frac{"d"y}{"d"x}#

#= (x + sin(y))frac{"d"y}{"d"x}#

#frac{"d"y}{"d"x} = -frac{y}{x + sin(y)}#

Cameron Alexander · Answer

undefined To find the implicit derivative of $1 = xy - \cos(y)$, differentiate both sides of the equation with respect to $x$ using the chain rule and implicit differentiation methods:

$$
\frac{d}{dx}(1) = \frac{d}{dx}(xy - \cos(y))
$$

Simplify:

$$
0 = y \frac{dx}{dx} + x \frac{dy}{dx} + \sin(y) \frac{dy}{dx}
$$

Solve for $\frac{dy}{dx}$:

$$
\frac{dy}{dx} = -\frac{y + x}{y + \sin(y)}
$$