# Homework Help: Differentiate this equation!

1. Jun 1, 2009

### vorcil

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

2: Ln(3x) / x^2

3: If dy/dx = 2ex 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 = 2ex 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

2. Jun 1, 2009

### rock.freak667

For 1) use the product law

$$\frac{d}{dx}(uv)=v\frac{du}{dx}+u\frac{dv}{dx}$$

2) Use the fact that d/dx(lnx)=1/x.

3) Integration is the reverse of differentiation. So d/dx(ex)=ex

4) Use your rules of logarithms here

5) Draw the line y=2x-1 and then draw x=1/2, then x=3/2. What figure do these lines and the x-axis form?

3. Jun 1, 2009

### vorcil

thanks i've retried

1: sin(4x^2) * 8x
sin(4x^2) * 8 + cos(4x^2)*12x*8x
8sin(4x^2) + 96x*cos(4x^2)
is that right?

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

4. Jun 1, 2009

### vorcil

3: If dy/dx = 2ex and y=6 when x=0, then y =

d/dx is also 2e^x + c = 6

2*e^x = 2, c=4
2e^x + 4 = 6
is this right?

5. Jun 1, 2009

### rock.freak667

Where did you get the 12x from in the second line of 1?
2 looks correct

ok if you know that when you differentiate ex with respect to x, you get ex+c.

and you differentiate y with respect to x to get ex, what do you get?

6. Jun 1, 2009

### vorcil

whoops should be 8x?
sin(4x^2)*8 + (cos(4x^2)*8x)*8x
8sin(4x^2)+(64x^2)*(cos(4x^2))

7. Jun 1, 2009

### vorcil

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

integral of 2x-1 = 2/2*x^2-x = (x^2)-x
((1.5^2)-1.5)-((0.5^2)-0.5) = 1

is that right?

8. Jun 2, 2009

### rock.freak667

I believe when they say in terms of an area, you should draw out what the integral represents. Draw y=2x-1,x=1/2,x=3/2 on the same graph and find the area of the enclosed figure.

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