How to differentiate this function f(x)=(x + 1)^3 (x − 3)^3?

f(x) = (x^2 - 2x - 3)^3 ---> f '(x) = 3(2x-2)(x^2-2x-3)^2

d [f(x)] = 3(x+1)^2 x 3(x-3)^2 (differentiation final answer)

1)y =(x+1)^3 into (x-^3

2) u =( x + 1 )^3

3) v =(x-3)^3 4)dy/dx =(v)du/dx+(u)dv/dx . 5)you have to treat x+1 as x and differentiate u ., then differentiate x+1 and multiply with t . Wait i'l show you how .if you differentiate x^3 you wil get 3x^2 , put x+1 instead of x in t . du/dx =3 (x+1)^2 .

6)similarly if u diff. V you 'll get

dv/dx =3(x-3)^2 .

7)now substitute du/dx and dv/dx in the formula .

8)dy/dx=[ (x-3)^3 (3)(x+1)^2 ] + [ ( x+1)^3 (3)(x-3)^2 ]

9)dy/dx=[ 3x^5 -21x^4+30x^3+216x^2+243x+81 ] + [3x^5-9x^4-12x^3 +36x^2 +72x+27]

10)dy/dx=(6x^5)-(30x^4)+(18x^3)+(252x^2)+(315x)+108 . Pshew . . . Its done . 27]

10)dy/dx=(6x^5)-(30x^4)+(18x^3)+(252x^2)+(315x)+108 . Pshew . . . Its done .

LET y= (x+1)^3 (x- 3)^3

dy/dx = (x--3)^3 d/dx(x+1)^3 + (x+1)^3 d/dx( x-3)^3

= (x-3)^3 3( x+1)^2 *d/dx(x+1)+ ( x+1)^3 3(x-3)^2 *d/dx(x-3)

= 3(x-3)^3(x+1)^2*1 + 3(x+1)^3 (x-3)^2 *1

= 3(x-3)^2 (x+1)^2 [ x-3 +x+1 ]

=3 (x-3)^2 (x+1)^2 [2x-2 ]

= 6 (x-3)^2(x+1)^2 (x-1) ans

Apply product rule first

ans=(x+1)^3 . d((x-3)^3)+ d((x+1)^3) . (x-3)^3 = (x+1)^3 .3(x-3)^2 . 1 + 3. (x+1)^2 . (x-3)^3

now expand and find out last result...

after differeniating the above eq the answer is

6(x^2-1)^5*2x

Take the derivative, set it equal to zero, and solve for x. Then take the second derivative, and plug in all solutions for x. Wherever f'' is less than zero, this is a local max. Wherever f'' is greater than zero, this is a local min. Wherever f'' is zero, this is a saddle point.

y = [(x + 1)^3] [(x − 3)^3]

Diff. w. r. t. x, we get,

d/dx(y) = d/dx { [(x + 1)^3] [(x − 3)^3] }

Therefore, dy/dx = { [(x + 1)^3] d/dx [(x − 3)^3] } + { [(x − 3)^3] d/dx [(x + 1)^3] }

dy/dx = { [(x + 1)^3](3)[(x − 3)^2](1) } + { [(x − 3)^3] 3 [(x + 1)^2] (1) }

dy/dx = (3) [(x+1)^2] [(x-3)^2] [ (x + 1) + (x - 3) ]

dy/dx = (3) [(x+1)^2] [(x-3)^2] [ (x - 2) ]

Final Answer :

dy/dx = (3) [(x+1)^2] [(x-3)^2] [ (x - 2) ]

Happy to help. :)