How do you differentiate # f(t)=-e^(sin(pi/x))sinpix # using the chain rule.?

Question

1971 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-10T22:30:50

#dy/dx={pi e^{sin(pi/x)} sin(pi x)cos(pi/x)}/{x^2}- pie^{sin(pi/x)} cos(pix)#

#y=-e^{sin(pi/x)}sin(pi x)#

#dy/dx=d/dx[-e^{sin(pi/x)}sin(pi x)]#

#dy/dx=d/dx[-e^{sin(pi/x)}]sin(pi x) + d/dx[ sin(pi x)](- e^{sin(pi/x)})#

#d/dx[ sin(pi x)](- e^{sin(pi/x)})= pi cos(pix)(- e^{sin(pi/x)})#

#d/dx[ sin(pi x)](- e^{sin(pi/x)})=- pie^{sin(pi/x)} cos(pix)#

Chain rule time: #alpha = sin(pi/x), beta = 1/x#

#d/dx[-e^{sin(pi/x)}] = d/{dalpha}[-e^{alpha}]d/{dbeta}[sin(pi beta)]d/dx[1/x]#

#d/dx[-e^{sin(pi/x)}] = -e^{alpha} times pi cos(pi beta) times (-1/x^2)#

#d/dx[-e^{sin(pi/x)}] = -e^{sin(pi/x)} times pi cos(pi/x) times (-1/x^2)#

#d/dx[-e^{sin(pi/x)}] = {pi e^{sin(pi/x)} cos(pi/x)}/{x^2}#

--

#d/dx[-e^{sin(pi/x)}]sin(pi x) = {pi e^{sin(pi/x)} sin(pi x)cos(pi/x)}/{x^2}#

Adding the two together we get:

#dy/dx={pi e^{sin(pi/x)} sin(pi x)cos(pi/x)}/{x^2}- pie^{sin(pi/x)} cos(pix)#