How do you differentiate #f(x)=1/sin(e^arctanx)# using the chain rule?

1 Answer
Nov 10, 2015

Answer:

#dy/dx = -(csc(e^(arctan(x)))cot(e^(arctan(x)))e^(arctan(x)))/(x^2+1)#

Explanation:

So we have

#y = 1/sin(e^(arctan(x))) = csc(e^(arctan(x)))#

While the cosecant has a tabled, standard derivative we can work with the sine too, if you don't remember / don't have a standard derivative table handy.

Let's say that #u = sin(e^(arctan(x)))#

#dy/dx = d/(du)u^(-1)(du)/dx#
#dy/dx = -u^(-3/2)d/dx(sin(e^(arctan(x)))#

Let's say that #v = e^(arctan(x))#

#dy/dx = -u^(-3/2)d/(dv)sin(v)(dv)/dx = -u^(-3/2)cos(v)d/dx(e^(arctan(x)))#

Let's say that #w = arctan(x)#

#dy/dx = -u^(-3/2)cos(v)d/(dw)e^(w)(dw)/dx#
#dy/dx = -u^(-3/2)cos(v)e^(w)d/dx(arctan(x))#
#dy/dx = -(u^(-3/2)cos(v)e^(w))/(x^2+1)#

Putting it all in terms of #x#

#dy/dx = -(csc(e^(arctan(x)))cot(e^(arctan(x)))e^(arctan(x)))/(x^2+1)#