Various Differentiation hang-ups.

C

Clauddvon

New member
Joined
Jan 29, 2006
Messages
5
Hi,

I'd like some help on a few differentiation problems that I'm stuck on. If anyone out there could help, I'd appreciate it.

1) Find the slope of the tangent line to the curve y = (sqrt(x+2)) at the point (0,2).

2) Find all values of x for which the curve y = x^3 + x^2 has a slope of 1.

3) The total distance traveled for a specific particle is given by s = t^2 + 3t + 4, where t is in seconds and s is in meters. The velocity (given in meters per second) of the particle at time t=2 is:

4) If f(x) = x^2 - 5x + 2 / 3x + 2, then f'(x) =

5) If f(x) = (x + 1)^2(x + 2)^3, then f'(x)=

6) If an object moves horizontally according to x = 2t + 1 and vertically according to y = 2x^3, find its vertical speed, dy/dt.
 
Clauddvon said:
Hi,

I'd like some help on a few differentiation problems that I'm stuck on. If anyone out there could help, I'd appreciate it.

1) Find the slope of the tangent line to the curve y = (sqrt(x+2)) at the point (0,2).

2) Find all values of x for which the curve y = x^3 + x^2 has a slope of 1.

3) The total distance traveled for a specific particle is given by s = t^2 + 3t + 4, where t is in seconds and s is in meters. The velocity (given in meters per second) of the particle at time t=2 is:

4) If f(x) = x^2 - 5x + 2 / 3x + 2, then f'(x) =

5) If f(x) = (x + 1)^2(x + 2)^3, then f'(x)=

6) If an object moves horizontally according to x = 2t + 1 and vertically according to y = 2x^3, find its vertical speed, dy/dt.
Click to expand...


1. Find the derivative (y') at x=0.
2. Find the y' and set it equal to 1.
3. Find the derivative at t=2.
4. First off, be sure to use parentheses correctly so that we know what you are asking. If the problem is f(x)=(x^2-5x+2)/(3x+2), use the quotient rule.
5. Use the product rule and power rule.
 
Please tell me if the two problems, below, have been worked correctly?

1) Find the slope of the tangent line to the curve y = (sqrt(x+2)) at the point (0,2).

dy / dx = 1/2(x + 2) ^-1/2
= 1 / 2(sqrt(x + 2))
= 1 / 2(sqrt(0 + 2))
= 1 / 2(sqrt(2))

3) The total distance traveled for a specific particle is given by s = t^2 + 3t + 4, where t is in seconds and s is in meters. The velocity (given in meters per second) of the particle at time t=2 is:

t = 2; s = t^2 + 3t + 4

s = (2)^2 + 3(2) + 4
s = 4 + 6 + 4
s = 14

v = ds / dt = d(14) / d(2) = 7 m/s
 
1. Correct.

3. You get the correct answer, but it was not quite solved properly.

First, we find s'=2t+3

s' at t=2=2(2)+3=7.
 
Okay, for this problem:

4) If f(x) = x^2 - 5x + 2 / 3x + 2, then f'(x)

Here is what I've tried to work and am not being too sucessful. Any help would be appreciated.

Okay,

let f(x) = x^2 - 5x + 2 and g(x) = 3x + 2, then:

f'(x) = g(x) f'(x) - f(x) g'(x) / [g(x)]^2
= ((3x + 2) d / dx(x^2 - 5x + 2) - (x^2 - 5x + 2) d / dx (3x + 2)) / (3x + 2)^2
= ((3x + 2) (2x - 5) - (x^2 - 5x + 2) (3)) / (3x + 2)^2

~The rest is what I believe I'm messing up on, but I just can't figure out where I'm going wrong.

= (6x^2 + 4x - 15x - 10 - 3x^2 + 15 - 6) / (3x + 2)^2
= 3x^2 - 11x - 1 / (3x + 2)^2
 
Clauddvon said:
Okay, for this problem:

4) If f(x) = x^2 - 5x + 2 / 3x + 2, then f'(x)

Here is what I've tried to work and am not being too sucessful. Any help would be appreciated.

Okay,

let f(x) = x^2 - 5x + 2 and g(x) = 3x + 2, then:

f'(x) = g(x) f'(x) - f(x) g'(x) / [g(x)]^2
= ((3x + 2) d / dx(x^2 - 5x + 2) - (x^2 - 5x + 2) d / dx (3x + 2)) / (3x + 2)^2
= ((3x + 2) (2x - 5) - (x^2 - 5x + 2) (3)) / (3x + 2)^2

~The rest is what I believe I'm messing up on, but I just can't figure out where I'm going wrong.

= (6x^2 + 4x - 15x - 10 - 3x^2 + 15 - 6) / (3x + 2)^2
^Where you wrote +15-6....There's your mistake....3(5x)=15x...you lost the x.

= 3x^2 - 11x - 1 / (3x + 2)^2
Click to expand...
 
Below is what I worked for Problem 5 and it's not working out:

5) If f(x) = (x + 1)^2(x + 2)^3, then f'(x)=

My work:

f(x) = (x+1)^2 (x+2)^3
= f'(x) g(x) + (f(x) g'(x)
= d / dx (x+1)^2 (x+2)^3 + (x+1)^2 d / dx (x+2)^3
= d / dx (2x+1)(3x+2) + (2x+1) d / dx (3x+2)
= (2x)(3x+2) + (2x+1)(3)
= 6x^2 + 10x + 4

If this is the correct answer, then I just need to figure out how to factor it correctly.
 
You must log in or register to reply here.
Top