Differenciate h(x) = x^3 (3x-5)^2
using power and product rule (also chain)

Question

Differenciate h(x) = x^3 (3x-5)^2 
using power and product rule (also chain)

campbell_st · Answer

use the product rule  

u = x^3     du/dx = 3x^2

$$v = (3x-5)^2$$

$$dv/dx = 2(3x-5) \times 3$$

dv/dx = 6(3x - 5)

$$dy/dx = du/dx \times v + dv/dx \times u$$

dy/dx = 3x^2 (3x-5)^2 + 6(3x-5) x^3

dy/dx = x^2(3x -5)(9x - 15 + 6x)
            = x^2(3x -5)(15x - 15)

anonymous · Answer

for 2(3x-5) (3) (x^3)
what i did is (6x^3 ) (3x-5)

so then when i expand it, the answer is different from urs..but i dont see any mistake doing this way...
what did i do wrong?

anonymous · Answer

and since the previous one has 3x^2 ..so i took out the common factor which is 3x^2

campbell_st · Answer

well I found x^2(3x-5) as common factor... then look at what was left

campbell_st · Answer

yep... you can do that..

anonymous · Answer

but i got 9x^2) ( 3x-5) (x-1) as my answer..

anonymous · Answer

（3x^2) (3x+5) [(3x-5 + 2)]

anonymous · Answer

aand then 
(3x^2) (3x+5) 3x-3 

which the answer is (9x^2) (3x+5)(x-1)

campbell_st · Answer

taking 3x^2(3x - 5) leaves  (3x - 5 + 2x)
            3x^2(3x -5)(5x - 5)

and you could now take 5 as a common factor from the last term

15x^2(5x - 3)(x -1)

campbell_st · Answer

its 2x and not 2.... x^3 = x^2 and x

anonymous · Answer

ohhhh!!!! thank you so much!

anonymous · Answer

campbell how u write the powers as we write in book without using ^