Y=(x^2-2x-3)^3Y=(x2−2x−3)3
Let, (x^2-2x-3)=t, so, Y=t^3, t=x^2-2x-3.(x2−2x−3)=t,so,Y=t3,t=x2−2x−3.
Thus, YY is a fun. of t,t, &, tt of x.x.
In such cases, (dY)/dxdYdx 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.!
