How Do You Simplify Derivatives in Basic Differentiation Problems?

[FaraDazed]

Homework Statement

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

Homework Equations

chain rule
product rule
quotient rule

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

 

[mtayab1994]

Homework Statement

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

Homework Equations

chain rule
product rule
quotient rule

The Attempt at a Solution

Part A:
$$y=0.2x^5 - sin4x + cos4x \\ y'=x^4-4cos4x-4sin4x$$

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

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

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.

 

[FaraDazed]

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.

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.

 

[Mark44]

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.

 

[FaraDazed]

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.

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.

 

[FaraDazed]

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

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

 

[mtayab1994]

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.

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.

 

[FaraDazed]

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.

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

like that? Thanks.

 

[mtayab1994]

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

like that? Thanks.

Yep that's it!

 

[FaraDazed]

Yep that's it!

Thank you :)

 

[mtayab1994]

Thank you :)

Not a problem.

 

[FaraDazed]

Yep that's it!

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?

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

 

[Mark44]

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?

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

Stop here. Expanding the denominator serves no purpose, and there is an error in your work.

$$y'=\frac{x^2(x^2 +3)}{x^4+2x^2+1} \\ y'=\frac{x^2(x^2 +3)}{x^2(x^2+2+1)} \\ y'=\frac{x^2+3}{x^2+2+1} \\ y'=\frac{x^2+3}{x^2+3} \\ y'=1 \\$$

No.
x⁴ + 2x² + 1 ≠ x²(x² + 2 + 1)