- #1

vorcil

- 398

- 0

^{2}) * 8x

2: Ln(3x) / x^

^{2}

3: If dy/dx = 2e

^{x}and y=6 when x=0, then y =

4: If ln(x^3) - ln(x) = 12, then x=e^(blank)

5: evaluate the integral by integrating it in terms of an area

(not sure how to use latex) integral from 1/2 -> 3/2 (2x-1)dx

my attempts =]

1: sin(4x^2)*8x

using the chainrule for the first bit of the equation

cos(4x^2)*8x

which is cos(4x^2)*8x*8

so 64x*cos(4x^2) is what i got

2: ln(3x) / x^2

using the quotient rule, lo*dhi - hi*dlo / lo^2

differentiating ln(3x) using chain rule i get 3(ln(3x)), x^2 = 2x

so quotient rule equation is

( (x^2)*3ln(3x) ) - (ln(3x) *2x) /(x^2)^2

simplifying

(x^2)(3ln(3x)) - ln(3x)*2x / x^8

(x^2)(2ln(3x))-2x / x^8

that's about as far as i got

3: If dy/dx = 2e

^{x}and y=6 when x=0, then y =

i tried to remember the old exponential differentiation rules,

if C=e^x then ln(c) = x

if c=ln(x) then e^x = x

not quite sure how to solve it from there :)

4: If ln(x^3) - ln(x) = 12, then x=e^(blank)

is this one of those exponential rules?

5: evaluate the integral by integrating it in terms of an area

(not sure how to use latex) integral from 1/2 -> 3/2 (2x-1)dx