How to Differentiate y= sin(3x)/x^2

[vorcil]

hiya, do you mind checking my awnsers please? i think i may have gotten them all wrong.

a: y= sin(3x)/x^2

b: (x^3) * (cos(4x))

c: If dy/dx =5/x and y=6 when x=0, then y=

d: if ln(x^3) - ln(x) = 4, then x=e^something

e: if dy/dx = 3e^x, and y = 7 when x =0, then y =

my attempts

a: y= sin(3x)/x^2
lo*dhi - hi*dlo/lo^2
((x^2)*(3cos(3x))) - ( (2x)*(sin(3x)) )/((x^2)^2)

 

[vorcil]

b: (x^3) * (cos(4x))

( (3x^2) * cos(4x) ) + ( (x^3) * 4(-sin(4x)) )

 

[vorcil]

c: If dy/dx =5/x and y=6 when x=0, then y=

integral of 5/x= 5ln(x) + c
5ln(x) + c = 6
ln(0)=1 so
5*1 + c = 6
c=1, 5ln(x) + 1 = y

 

[vorcil]

d: if ln(x^3) - ln(x) = 4, then x=e^something

ln(x³)-ln(x) = 4

e^{ln(x^3)} - e^ln(x) = e⁴
= x³ - x = e⁴

not sure how to solve from here, mind helpin or showing me an example?

 

[rock.freak667]

a: y= sin(3x)/x^2
lo*dhi - hi*dlo/lo^2
((x^2)*(3cos(3x))) - ( (2x)*(sin(3x)) )/((x^2)^2)

This looks good

b: (x^3) * (cos(4x))

( (3x^2) * cos(4x) ) + ( (x^3) * 4(-sin(4x)) )

Good.

c: If dy/dx =5/x and y=6 when x=0, then y=

integral of 5/x= 5ln(x) + c
5ln(x) + c = 6
ln(0)=1 so
5*1 + c = 6
c=1, 5ln(x) + 1 = y

Are you sure those are the initial conditions? Since ln(0) does not exist.

d: if ln(x^3) - ln(x) = 4, then x=e^something

Use this logarithm rule $log_ax - log_a y=log_a (\frac{x}{y})$

 

[vorcil]

Are you sure those are the initial conditions? Since ln(0) does not exist.

x=e not 0, sorry

 

[rock.freak667]

x=e not 0, sorry

Then part c would be correct.

 

[vorcil]

Then part c would be correct.

so ln(e) = 1?

 

[vorcil]

d: if ln(x^3) - ln(x) = 4, then x=e^something

i'm not sure how to use that equation you gave me to solve this
$log_ax - log_a y=log_a (\frac{x}{y})$

 

[rock.freak667]

so ln(e) = 1?

Yes

d: if ln(x^3) - ln(x) = 4, then x=e^something

i'm not sure how to use that equation you gave me to solve this
$log_ax - log_a y=log_a (\frac{x}{y})$

If I tell you that lnX is the same as log_eX, would that help you more?

 

[vorcil]

d: if ln(x^3) - ln(x) = 4, then x=e^something
log_e^{x^3} - log_e^x

which is log_e^{(x^3/x)}
do i use the quotient rule or something there?

 

[vorcil]

e: if dy/dx = 3e^x, and y = 7 when x =0, then y =

integral of 3e^x dx = 3e^x + c
so 3e^x + c = 7
e^0 = 1,
3*1 + c = 7
y=3e^x + 4

 

[rock.freak667]

using this $log_aX - log_a Y=log_a (\frac{X}{Y})$

if X=x³ and Y=x, can you find one log term?

 

[vorcil]

using this $log_aX - log_a Y=log_a (\frac{X}{Y})$

if X=x³ and Y=x, can you find one log term?

is it x^2? x^3/x^1 = X^3-1

ln(x^2) = 4
x^2 = e^4
x = e^4/nothing^2
x = e^4-2

x=e^2?

 

[rock.freak667]

is it x^2? x^3/x^1 = X^3-1

ln(x^2) = 4
x^2 = e^4
x = e^4/nothing^2
x = e^4-2

x=e^2?

Yes, the final answer is correct, but it should be something like this

$$x^2=e^4$$
$$x=\pm (e^4)\frac{1}{2}$$

$$x= \pm e^2$$

But x can't be negative since ln(-ve) does not exist so

x=e² only.

 

[vorcil]

Yes, the final answer is correct, but it should be something like this

$$x^2=e^4$$
$$x=\pm (e^4)\frac{1}{2}$$

$$x= \pm e^2$$

But x can't be negative since ln(-ve) does not exist so

x=e² only.

OHO sweet! cheers mate, thanks for the help =]