Calculus - Differentiation and simplification?

Question

Differentiate the following and simplify as far as possible

Y= x^2 (2x - 3)^3

and

Y= [e^(3t)] / t+1

Answer 1

Question 1(Product rule)
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 (Quotient rule)
y = e^(3t) / (t + 1)
dy/dt = [ (t + 1).3e^(3t) - e^(3t).1] / (t + 1)²
dy/dt = e^(3t)[ 3.(t + 1) - 1 ] / (t + 1)²
dy/dt = e^(3t).[ 3t + 2 ] / (t + 1)²

Answer 2

the first one is
=(x^2)[6(2x-3)^2]+(2x-3)3(2x)
you use the product rule and you have to use chain rule for
(2x+3)^3
The second one is [e^3t(3t+2)] / [(t+1)^2]
you have to use quotient rule

Answer 3

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

Use the product rule to obtain

y' = 2x (2x - 3)^3 + (x^2)( 3(2x - 3)^2 (2) )
y' = 2x (2x - 3)^3 + 6x^2 (2x - 3)^2

Factor,

y' = 2x(2x - 3)^2 ( (2x - 3) + 3x )
y' = 2x(2x - 3)^2 ( 5x - 3 )

Answer 4

I hope u want differentiation w.r.t. x
i) dy/dx = x (7x-6) (2x-3)^2
ii)dy/dx =3e^(3t) / t+1

Answer 5

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

y' = (x^2)'(2x - 3)^3 + (x^2)((2x - 3)^3)'
y' = (2x)(2x - 3)^3 + 6(x^2)(2x - 3)^2
y' = (2x)(2x - 3)^2 * ((2x - 3) + 6x)
y' = (2x)(2x - 3)^2 * (8x - 3)
y' = (16x^2 - 6x)(2x - 3)^2
y' = (16x^2 - 6x)(4x^2 - 12x + 9)
y' = 64x^4 - 192x^3 + 144x^2 - 24x^3 + 72x^2 - 54x
y' = 64x^4 - 216x^3 + 216x^2 - 54x

ANS : y' = 64x^4 - 216x^3 + 216x^2 - 54x

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y = (e^(3t))/(t + 1)

y' = ((e^(3t)' * (t + 1)) - ((e^(3t))(t + 1)'))/((t + 1)^2)
y' = ((e^(3t) * (3t)')(t + 1) - e^(3t))/((t + 1)^2)
y' = (e^(3t)(3't + 3t')(t + 1) - e^(3t))/((t + 1)^2)
y' = (3e^(3t)(t + 1) - e^(3t))/((t + 1)^2)
y' = ((e^(3t))(3t + 3 - 1))/((t + 1)^2)
y' = ((3t + 2)(e^(3t)))/((t + 1)^2)
y' = (3te^(3t) + 2e^(3t))/((t + 1)^2)