# Homework Help: Basic Differentiation

1. May 14, 2013

1. The problem statement, all variables and given/known data
Differentiate the following and display in the simplest form.

Part A:
$$y=0.2x^5 - sin4x + cos4x$$

Part B:
$$y=(2x^4 +3)^3$$

Part C:
$$y=2x^3 sinx$$

Part D:
$$y=\frac{sinx}{x^2}$$

Part E:
$$y=\frac{x^3}{x^2 +1}$$

2. Relevant equations
chain rule
product rule
quotient rule

3. The attempt at a solution
Part A:
$$y=0.2x^5 - sin4x + cos4x \\ y'=x^4-4 \ cos4x-4 \ sin4x$$

Part B:
$$y=(2x^4 +3)^3 \\ let \ u =2x^4 +3 \\ u'=8x^3 \\ y'(u)=3u^2 \\ y'=3u^2 \times 8x^3 \\ y'=3(2x^4 +3)^2 \times 8x^3 \\ y'=24x^3(2x^4+3)^2 \\$$

Part C:
$$y=2x^3 \ sinx \\ let \ u = 2x^3 \\ let \ v = sinx \\ y'=vu'+uv' \\ y'=sinx \ 6x^2 + 2x^3 \ cosx \\ y'=2x^2(3 \ sinx + x \ cosx)$$

Part D:
$$y=\frac{sinx}{x^2} \\ y'=\frac{vu'-uv'}{v^2} \\ y'=\frac{x^2 cosx - sinx \ 2x}{(x^2)^2} \\ y'=\frac{x(x \ cosx - 2sinx)}{x^4} \\ y'=x^{-3}(x \ cosx - 2sinx)$$

Part E:
$$y=\frac{x^3}{x^2 +1} \\ y'=\frac{vu'-uv'}{v^2} \\ y'=\frac{(x^2 +1)3x^2-x^32x}{(x^2 + 1)^2} \\ y'=\frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} \\ y'=\frac{x^2(3x^2 + 3 - 2x^2 )}{(x^2 +1)^2} \\$$

Last edited: May 14, 2013
2. May 14, 2013

### mtayab1994

A, B and C are ok . D is ok too just cancel that x you factored out with one x from the denominator to get it in simplest form. E is wrong try to rework it.

3. May 14, 2013

Thanks for taking a look :), I have edited part D, is that what you meant?

On Part E, is it my jump from line 3 to line 4 where I went wrong or was it before that?

Thanks.

4. May 14, 2013

### Staff: Mentor

In the next to last step in part E, simplify the numerator first by combining like terms.

Also, it is best to post only a single problem, or two at most, in a thread.

5. May 14, 2013

Here is my rework of Part E, I went around it a different way by expanding the denomator and then cancelling terms:
Part E:
$$y=\frac{x^3}{x^2 +1} \\ y'=\frac{vu'-uv'}{v^2} \\ y'=\frac{(x^2 +1)3x^2-x^32x}{(x^2 + 1)^2} \\ y'=\frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} \\ y'=\frac{x^4 + 3x^2}{x^4 + 2x^2 +1} \\ y'=\frac{3x^2}{2x^2+1}$$

Does this work?

Thanks.

Last edited: May 14, 2013
6. May 14, 2013

OK no problem, will do, I just didn't want to create 3+ threads all in one go.

7. May 14, 2013

### mtayab1994

No that does not work you can't just cancel out an x^4 because you haven't factored with it.

You third to last line is correct you will get x^4+3x^2 and just factor that out and you are done.

8. May 14, 2013

$$y=\frac{x^3}{x^2 +1} \\ y'=\frac{vu'-uv'}{v^2} \\ y'=\frac{(x^2 +1)3x^2-x^32x}{(x^2 + 1)^2} \\ y'=\frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} \\ y'=\frac{x^4 + 3x^2}{(x^2 + 1)^2} \\ y'=\frac{x^2(x^2 +3)}{(x^2 + 1)^2} \\$$

like that? Thanks.

9. May 14, 2013

### mtayab1994

Yep that's it!

10. May 14, 2013

Thank you :)

11. May 14, 2013

### mtayab1994

Not a problem.

12. May 22, 2013

Hi again!

I have just been playing around with part E again and think I may have made it even simpler, can you tell me if what I have done below is OK?

$$y=\frac{x^3}{x^2 +1} \\ y'=\frac{vu'-uv'}{v^2} \\ y'=\frac{(x^2 +1)3x^2-x^32x}{(x^2 + 1)^2} \\ y'=\frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} \\ y'=\frac{x^4 + 3x^2}{(x^2 + 1)^2} \\ y'=\frac{x^2(x^2 +3)}{(x^2 + 1)^2} \\ y'=\frac{x^2(x^2 +3)}{x^4+2x^2+1} \\ y'=\frac{x^2(x^2 +3)}{x^2(x^2+2+1)} \\ y'=\frac{x^2+3}{x^2+2+1} \\ y'=\frac{x^2+3}{x^2+3} \\ y'=1 \\$$

13. May 22, 2013

### Staff: Mentor

Stop here. Expanding the denominator serves no purpose, and there is an error in your work.
No.
x4 + 2x2 + 1 ≠ x2(x2 + 2 + 1)