Calculus Differentiation - simplifying?

Question

1.y=x^2(2x-3)^3

2.y= e^3^t/t+1

How do i differentiate the above simplified as much/far as possible

y= e^3^t (e with 3 and t as superscript bside)
-------
t+1

Answer 1

1)

product rule:

y = f(x) * g(x)
y' = f'g + fg'

so...
y = x^2 (2x - 3)^3
f(x) = x^2 ==> f'(x) = 2x
g(x) = (2x - 3)^3 ==> g'(x) = 3(2x - 3)^2 (2) = 6(2x - 3)^2

so...
y' = (2x)(2x - 3)^3 + (x^2)(6)(2x - 3)^2
===================================
2)
y = e^3t / (t + 1)

Use the quotient rule.

Derivative of (f / g) = (f'g - fg') / g^2

f(x) = e^3t
so... f ' (x) = 3(e^3t)

g(x) = t + 1
so... g ' (x) = 1

y' = [3(e^3t)(t + 1) - 1(e^3t)] / (t + 1)^2

Answer 2

1. y = (x^2)*((2x-3)^3)
y' = u'*v + v'*u (u and v are functions of x)

u = x^2
u' = 2x
v = (2x-3)^3
v' = (3*(2x-3)^2)*2

y' = [(2x)*((2x-3)^3)] + [((3*(2x-3)^2)*2)*(x^2)]
y' = [(2x-3)^2]*[10(x^2)-6x] (Factored version)
y' = 40x^4 - 144x^3 + 162x^2 - 54x

y'' = 160x^3 - 432x^2 + 324x - 54

y''' = 480x^2 - 864x + 324

y'''' = 960x - 864

y''''' = 960

You can differentiate number one 5 times.
I don't understand #2, sorry!

Answer 3

Question 1
y = x².(2x - 3)³
dy/dx = 2x.(2x - 3)³ + 3.(2x - 3)².(2).(x²)
dy/dx = 2x.(2x - 3)².[(2x - 3) + 3x ]
dy/dx = 2x.(2x - 3)².(5x - 3)
Question 2
Guessing that questiion is:-
y = e^(3t) / (t + 1)
dy/dt = [(t + 1).3e^(3t) + e^(3t)] / (t + 1)²
dy/dt = e^(3t).[3t + 3 + 1] / (t + 1)²
dy/dt = e^(3t).[ (3t + 4) / (t + 1)² ]

Answer 4

que is not clear
what we hav to find dy/dx or dx/dy or dy/dt?

Answer 5

y=x^2/(2x-3)^3
dy/dx=2x/()2x-3)^3-6x^2/(2x-3)^4answer