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

1 Answer
Apr 10, 2018

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

Explanation:

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)