Find #dy/dx# for the function #e^(cos z) + e^(sin y) =1/4 #?
2 Answers
Feb 22, 2018
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))#
Feb 23, 2018
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