D (log sin cos tan e^cosx)/dx ?

1 Answer
Jul 9, 2018

Answer:

#= cot(cos( tan e^cosx)) * sin(tan (e^cosx)) * sec^2 ( e^cosx) * e^cosx * sinx#

Explanation:

The chain rule for a single composition is:

  • #(f_1 circ f_2)' = f_1' f_2'#

The Lagrange/ prime notation of the RHS signifies differentiation with respect to the function's own argument.

Extending for multiple composition:

  • #(f_1 circ f_2 circf_3....)' = f_1' f_2'f_3'...#

For: # log (sin (cos (tan (e^cosx)))))#

  • Let: # {(a_1(x) = sin (cos (tan (e^cosx)))),(a_2(x) = cos( tan (e^cosx))),(a_3(x) = tan (e^cosx ) ), (a_4(x) = e^cosx ),(a_5(x) = cosx ) :} #

#implies d/dx ( log (sin (cos (tan( e^cosx))))) qquad square#

#= underbrace(1/ a_1)_(= (log a_1)^') * cosa_2 * (- sina_3) * sec^2 a_4 * a_4 * underbrace(-sinx)_(=a_5')#

#= 1/ (sin (cos (tan (e^cosx)))) * cos(cos( tan( e^cosx))) * sin(tan (e^cosx)) * sec^2 ( e^cosx) * e^cosx * sinx#

#= cot(cos( tan( e^cosx))) * sin(tan ((e^cosx))) * sec^2 ( e^cosx) * e^cosx * sinx#

In Liebnitz notation:

  • #square = (d(loga_1))/(d a_1)(d a_1)/(d a_2)(d a_2)/(d a_3)(d a_3)/(d a_4)(d a_4)/(d a_5) (d a_5)/dx bb( = (d(loga_1))/(d x))#