Find #dy/dx# for the function #e^(cos z) + e^(sin y) =1/4 #?

Question

Aurora Alvarado · Accepted Answer

#"see explanation"# #"differentiate "color(blue)"implicitly with respect to x"# #rArre^(cosx)(-sinx)+e^(siny)(cosydy/dx)=0# #rArr-sinxe^(cosx)+cosye^(siny)dy/dx=0# #rArrdy/dx(cosye^(siny))=sinxe^(cosx)# #rArrdy/dx=(sinxe^(cosx))/(cosye^(siny))#

Aurora Alvarado · Answer

We can apply the Implicit Function Theorem: If we define: # f(x,y) = e^(cos z) + e^(sin y) -1/4 # Then: # dy/dx = -f_x/f_y # And we calculate the partial derivatives: # f_x = (partial f)/(partial x) #
# \ \ \  = e^(cosx)(-sinx) # # f_y = (partial f)/(partial y) #
# \ \ \  = e^(siny)(cosy) # So then: # dy/dx = - (e^(cosx)(-sinx)) / (e^(sinx)(cosx)) #
# \ \ \ \ \ \ = (sinx \ e^(cosx)) / (cos y \ e^(siny)) \ \ \ # QED

Scarlett Allison · Answer

undefined To find $ \frac{dy}{dx} $ for the function $ e^{\cos(z)} + e^{\sin(y)} = \frac{1}{4} $:

1. Differentiate both sides of the equation with respect to $ x $.
2. Apply the chain rule and the derivative of $ e^u $, where $ u $ is a function of $ x $, to each term.
3. Solve for $ \frac{dy}{dx} $.

The steps are as follows:

$ \frac{d}{dx}(e^{\cos(z)}) + \frac{d}{dx}(e^{\sin(y)}) = \frac{d}{dx}\left(\frac{1}{4}\right) $

Using the chain rule and derivative of $ e^u $:

$ -\sin(z) \cdot \frac{dz}{dx} \cdot e^{\cos(z)} + \cos(y) \cdot \frac{dy}{dx} \cdot e^{\sin(y)} = 0 $

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

$ \frac{dy}{dx} = \frac{\sin(z) \cdot \frac{dz}{dx} \cdot e^{\cos(z)}}{\cos(y) \cdot e^{\sin(y)}} $