#Y=(x^2-2x-3)^3#
Let, #(x^2-2x-3)=t, so, Y=t^3, t=x^2-2x-3.#
Thus, #Y# is a fun. of #t,# &, #t# of #x.#
In such cases, #(dY)/dx# can be obtained by The Chain Rule :-
#"The Chain Rule : "(dY)/dx=((dY)/dt)((dt)/dx)............(star).#
Recall that, #d/dt(t^n)=nt^(n-1).#
Hence, #Y=t^3 rArr (dY)/dt=3t^2...(1), and, t=x^2-2x-3#
#rArr dx/dt=2x-2=2(x-1).............(2).#
Using #(1) and (2) in (star),# we get,
#(dY)/dx=(3t^2){2(x-1)}=6(x-1)t^2,# &, returning back from #t# to #x#,
#(dY)/dx=6(x-1)(x^2-2x-3)^2=6(x-1){(x-3)(x+1)}^2, or,#
#(dY)/dx=6(x-1)(x-3)^2(x+1)^2.#
Regarding, the Diffn. of #y=sin2xcos3x,# we have, #2# Methods :-
Method 1:-
We know, that, #2cosAsinB=sin(A+B)-sin(A-B).#
With, #A=3x, B=2x, 2cos3xsin2x=sin(3x+2x)-sin(3x-2x)#
#:. y=cos3xsin2x=1/2{sin5x-sinx}.#
Therefore, using the Chain Rule,
#dy/dx=1/2{(cos5x)d/dx(5x)-cosx}, or, #
# dy/dx=1/2{5cos5x-cosx}.#
Method 2:-
In this Method, we use the following Product Rule, together
with the Chain Rule.
#"Product Rule : "d/dx(uv)=u(dv)/dx+v(du)/dx.#
#:. dy/dx=(cos3x)d/dx(sin2x)+(sin2x)d/dx(cos3x)#
#=(cos3x){(cos2x)d/dx(2x)}+(sin2x){(-sin3x)d/dx(3x)}#
#:. dy/dx=2cos3xcos2x-3sin3xsin2x.#
I leave it upon the reader to show that both the Answers are same.
Enjoy Maths.!
