Question #30f2b

1 Answer
Mar 7, 2017

Answer:

# (1) : dy/dx=1/2{5cos5x-cosx}, or, (1)dy/dx=2cos3xcos2x-3sin3xsin2x.#

# (2) : (dY)/dx=6(x-1)(x-3)^2(x+1)^2.#

Explanation:

#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.!