What are the derivatives?! Please help!!?

Question

a)y= ln (x^2+3x+4)
b)y= log x^2
c)y= ln x^2/ (3x-4)^3
d)y= 2x^3 e^4x
e)y= Square root[x^3+e^-x + 5]

Answer 1

a) y=ln(x²+3x+4)
dy/dx = (2x+3)/(x²+3x+4)

b)y=logx²
dy/dx = 2x/x²

c)y=lnx² / (3x-4)³
dy/dx = (3x-4)³(2x/x²) - (lnx²)(3(3x-4)².(3)) / (3x-4)^6
dy/dx =( (3x-4)³(2/x) - 9(lnx²)(3x-4)² ) / ( 3x-4)^6

d)y= 2x³e^4x
dy/dx = 2x³(e^4x)(4) + (2x²)(e^4x)
dy/dx = (e^4x)(8x³+2x²)

e)y=√(x³+e^-x+5)
dy/dx = ½(x³+e^-x +5)^-½ . (3x²+e^-x(-1) )
dy/dx = (3x²-e^-x) / 2√(x³+e^-x +5)

Answer 2

a)y= ln (x^2+3x+4)
dy/dx = (2x + 3)/(x^2+3x+4) ANS

b)y= log x^2
y = 2logx
dy/dx = 2log(e)/x ANS

where log(e) = M = 0.43429

c)y= ln x^2/ (3x-4)^3
y = lnx^2 - ln(3x-4)^3
y = 2lnx - 3ln(3x-4)
dy/dx = 2log(e)/x - 3(log(e))(3)/(3x-4)
dy/dx = log(e)[2/x - 9/(3x - 4)]

d)y= 2x^3 e^4x
dy/dx = 2[3x^2(e^4x) + x^3(e^4x)(4)]
dy/dx = 2[3x^2(e^4x) + 4x^3(e^4x)]
dy/dx = 2x^2(e^4x)[3 + 4x] ANS

e)y= Square root[x^3+e^-x + 5]
y = (x^3 + e^-x + 5)^(1/2)
dy/dx = {(1/2)(x^3 + e^-x + 5)^(-1/2)}[3x^2 + (e^-x)(-1)]
dy/dx ={(1/2)(x^3 + e^-x +5)^(-1/2)}[3x^2 - e^-x] ANS

teddy boy

Answer 3

(2x+3)/(x^2+3x+4)

2x/x^2 = 2/x

(3x-4)^3/x^2 * [2x(3x-4)^3 -6x(3x-4^2]/(3x-4)^6
= (3x-4)^3/x^2 [2x(3x-4) -6x]/(3x-4)^4
= 2(3x-7)/(x(3x-4))

= 8x^3e^4x + 3x^2e^4x = (8x^3+3x^2)e^4x

=(3x^2-e-x)/[2sqrt(x^3 +e-x+5)]