using the product rule twice simplification

[RafN]

i'm doing exercises on Khan academy and i'm struggeling with simplifications when it comes to taking the derivative of three terms that require using the productrule twice. The first step is not te problem, it's the simplification that i'm struggeling to grasp.

For exemple: find the derivative of 2x*x^1/2*sin(x)

i get that taking the derivative twice would result in something that looks like the following:
u′(x)⋅v(x)⋅w(x)+u(x)⋅v′(x)⋅w(x)+u(x)⋅v(x)⋅w′(x)

and i got this step down with this step i got



But the next step, simplification, i can't follow and feel they skip too many steps in between for me to comprehend:

 

[Deleted member 4993]

i'm doing exercises on Khan academy and i'm struggeling with simplifications when it comes to taking the derivative of three terms that require using the productrule twice. The first step is not te problem, it's the simplification that i'm struggeling to grasp.

For exemple: find the derivative of 2x*x^1/2*sin(x)

i get that taking the derivative twice would result in something that looks like the following:
u′(x)⋅v(x)⋅w(x)+u(x)⋅v′(x)⋅w(x)+u(x)⋅v(x)⋅w′(x)

and i got this step down with this step i got

View attachment 34239

But the next step, simplification, i can't follow and feel they skip too many steps in between for me to comprehend:

View attachment 34240

When you tried to simplify -

What did you get ? Please share your work.

 

[RafN]

This is the furthest i got, but frankly i don’t think i’m anywhere near correct.

 

[JeffM]

We have

[math] y’ = 2 * \sqrt{x} * \sin (x) + 2x * \dfrac{1}{2\sqrt{x}} * \sin (x) + 2x * \sqrt{x} * \cos (x) = \\ 2 * \sqrt{x} * \sin (x) + \dfrac{2(\sqrt{x})^2}{2 \sqrt{x}} * \sin (x) + 2x * \sqrt{x} * \cos (x) =\\ 2 * \sqrt{x} * \sin (x) + \sqrt{x} * \sin (x) + 2x * \sqrt{x} * \cos (x) = \\ \sqrt{x} * \{2 \sin (x) + \sin (x) + 2x \cos (x) \} \implies\\ y’ = \sqrt{x} \{3 \sin (x) + 2x \cos (x) \}. [/math]

 

[blamocur]

Can you simplify [imath]x \cdot x^{1/2}[/imath] ?

 

[Steven G]

2 + 1/2 + 4. Only the 1 is being divided by 2. The factors 2 and 4 are NOT being divided by 2. You made a similar mistake in your work.

 

[RafN]

We have

[math] y’ = 2 * \sqrt{x} * \sin (x) + 2x * \dfrac{1}{2\sqrt{x}} * \sin (x) + 2x * \sqrt{x} * \cos (x) = \\ 2 * \sqrt{x} * \sin (x) + \dfrac{2(\sqrt{x})^2}{2 \sqrt{x}} * \sin (x) + 2x * \sqrt{x} * \cos (x) =\\ 2 * \sqrt{x} * \sin (x) + \sqrt{x} * \sin (x) + 2x * \sqrt{x} * \cos (x) = \\ \sqrt{x} * \{2 \sin (x) + \sin (x) + 2x \cos (x) \} \implies\\ y’ = \sqrt{x} \{3 \sin (x) + 2x \cos (x) \}. [/math]
2 * \sqrt{x} * \sin (x) + \dfrac{2(\sqrt{x})^2}{2 \sqrt{x}} * \sin (x) + 2x * \sqrt{x} * \cos (x) =\\

i never thought about rewriting 2x over 2 squareroot(x) as 2 squareroot of (x)^2 and then canceling out the denominator. thank you so much!

 

[RafN]

Can you simplify [imath]x \cdot x^{1/2}[/imath] ?

x^(3/2) ?

 

[JeffM]

i never thought about rewriting 2x over 2 squareroot(x) as 2 squareroot of (x)^2 and then canceling out the denominator. thank you so much!

Couple of minor points.

You are being careless with grouping symbols. x^3/2 means [imath]\dfrac{x^3}{2}[/imath] rather than [imath]x^{3/2}[/imath].

I do not know whether they continue to teach rationalizing the denominator. If they do, you can get to the same place this way

[math]2x * \dfrac{1}{2\sqrt{x}} * \sin (x) = \cancel 2x * \dfrac{1}{\cancel 2 \sqrt{x}} * \dfrac{\sqrt{2}}{\sqrt{x}} \sin (x) =\\ x * \dfrac{\sqrt{x}}{x} * \sin (x) = \sqrt{x} * \sin (x).[/math]
Blamocur was giving a third (more systematic) way to simplify: use exponents, simplify, and look for a common factor. The first term then has [imath]x^{1/2}[/imath] as a factor; the second term has [imath]x^1 * x^{-1/2}[/imath] as factors, and the third term has [imath]x^1 * x^{1/2}[/imath] as factors. Using the laws of exponents, we end up with [imath]x^{1/2}[/imath] as a factor for the first and second term and [imath]x^{3/2}[/imath] as a factor for the third term. It is then easy to see that [imath]x^{1/2} = \sqrt{2}[/imath] is a common factor in all three terms.

In short, there are at least three different ways to simplify that derivative.