Differentiation with fractions, radicands, and the power chain rule

In summary: This matches the book's answer. :DIn summary, we learned how to differentiate two different problems using the quotient rule and the power/chain rule. The first problem involved differentiating x divided by the square root of x squared plus 1, which resulted in a simplified form of (x squared plus 1) to the negative one-half minus x squared times (x squared plus 1) to the negative three-halves. The second problem involved differentiating the square root of x plus 2 divided by the square root of x minus 1, which resulted in a simplified form of negative three divided by two times the square root of the quantity x plus 2 times the quantity x minus 1 cubed.
  • #1
SHLOMOLOGIC
4
0
Differentiate the following two problems.

1. x divided by the square root of x squared+ 1

2. The square root of x + 2
divided by the square root of x - 1

Thank you.
 
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  • #2
Hello and welcome to MHB, SHLOMOLOGIC! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?
 
  • #3
MarkFL said:
Hello and welcome to MHB, SHLOMOLOGIC! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?
Find derivative of X/sqrt (x^2 + 1))

using the quotient rule and the power chain rule

x/(x^2 + 1) ^1/2

derivative of x=1
derivative of denominator=1/2(x^2 +1)^-1/2 times 2x

quotient rule:

[( x^2 + 1)^1/2 times 1] ( minus) x times [above derivative of denominator]

all of the above line divided by the the denominator squared

I get (x^2+1)^1/2 (minus) x times ½ (x^2 + 1) ^-1/2 (times) 2x/ x^2 + 1

I then reduce this to (x^2 + 1)^-1/2 –- x^2(x^2 + 1)^--3/2

This does not match the book answer, 1/(x^2 + 1)^ 3/2
 
  • #4
Okay, we are given:

\(\displaystyle f(x)=\frac{x}{\sqrt{x^2+1}}=x\left(x^2+1\right)^{-\frac{1}{2}}\)

If we apply the product rule, we find:

\(\displaystyle f'(x)=x\left(-\frac{1}{2}\left(x^2+1\right)^{-\frac{3}{2}}2x\right)+(1)\left(x^2+1\right)^{-\frac{1}{2}}=\left(x^2+1\right)^{-\frac{1}{2}}-x^2\left(x^2+1\right)^{-\frac{3}{2}}\)

This is what you got using the quotient rule. :) Now if we factor, we obtain:

\(\displaystyle f'(x)=\left(x^2+1\right)^{-\frac{3}{2}}\left(x^2+1-x^2\right)=\left(x^2+1\right)^{-\frac{3}{2}}\)
 
  • #5
For the second problem, I think I would write:

\(\displaystyle f(x)=\left(\frac{x+2}{x-1}\right)^{\frac{1}{2}}\)

Then, I would apply the power/chain/quotient rules. Can you proceed?
 
  • #6
MarkFL said:
For the second problem, I think I would write:

\(\displaystyle f(x)=\left(\frac{x+2}{x-1}\right)^{\frac{1}{2}}\)

Then, I would apply the power/chain/quotient rules. Can you proceed?

Thanks for responding.

First, where did you get the square root symbols and the textbook notation?

Secondly, I did try your method. I believe the idea is to, step one, use the power rule on the entire expression. Step two involves using the quotient rule to determine the derivative of the fraction. Then multiply and factor. factor.

Perhaps I made an error in my factoring and/or my square-root arithmetic; I'll certainly try it again, but I'm worn out after several hours of failed attempts yesterday.

I don't believe my result can be transformed into the book answer. I ended up with (x+2) to the one-half/(x-1) to the five-halfs--or is it halves?
 
  • #7
Got it. Finally got my arithmetic right. It's easy when you know how.
 
  • #8
To make the application of the chain rule a bit simpler, we can utilize:

\(\displaystyle \frac{x+2}{x-1}=1+3(x-1)^{-1}\)

And so:

\(\displaystyle f'(x)=\frac{1}{2}\left(\frac{x+2}{x-1}\right)^{-\frac{1}{2}}\left(-3(x-1)^{-2}\right)=-\frac{3}{2}\left(\frac{x-1}{x+2}\right)^{\frac{1}{2}}(x-1)^{-2}=-\frac{3}{2\sqrt{(x+2)(x-1)^3}}\)
 

FAQ: Differentiation with fractions, radicands, and the power chain rule

What is differentiation?

Differentiation is a mathematical concept that involves finding the rate of change of a function with respect to its independent variable. It is used to determine the slope of a curve at a specific point, and is an essential tool in calculus and other areas of mathematics.

How does differentiation work with fractions?

When differentiating a fraction, the power rule can be applied to both the numerator and denominator separately. The quotient rule can also be used in some cases. It is important to simplify the expression before differentiating to make the process easier.

What is a radicand?

A radicand refers to the number or expression under a radical symbol, such as a square root or cube root. In the context of differentiation, the radicand can be treated as a variable and the power rule can be applied to find the derivative.

How do you apply the power chain rule in differentiation?

The power chain rule is used when differentiating functions that involve nested exponents, such as (x^2)^3. To apply the rule, the derivative of the outer function is multiplied by the derivative of the inner function, which is then multiplied by the derivative of the original function.

What are some common mistakes when differentiating with fractions, radicands, and the power chain rule?

Some common mistakes include not simplifying the expression before differentiating, forgetting to apply the power rule to both the numerator and denominator in fractions, and not properly applying the power chain rule to nested exponents. It is important to carefully follow the rules of differentiation and simplify the expression as much as possible before calculating the derivative.

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