D (log sin cos tan e^cosx)/dx ?
1 Answer
Jul 9, 2018
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:
- 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 ) :} #
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))#