# Differentiation with square roots

• MHB
• Yankel
In summary, the two functions have different derivatives depending on the order of the rationalizations.
Yankel
Hello all,

I was trying to find derivatives of two functions containing square roots. I got answers which I believe should be correct, however, the answers in the book differ significantly. The first answer of mine was checked in MAPLE and found correct. My guess that the author made some algebraic manipulations but I can't seem to track it down and get to the same result. Can you kindly take a look ?

The functions are:

$f(x)=(\sqrt{x}+\frac{1}{\sqrt{x}})^{10}$

$g(x)=\frac{\sqrt{x}}{\sqrt{x}-\sqrt{x-1}}$

$f'(x)=5\cdot (\sqrt{x}+\frac{1}{\sqrt{x}})^{9}\cdot (\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x^{3}}})$

$g'(x)=\frac{\frac{1}{2\sqrt{x}}\cdot (\sqrt{x}-\sqrt{x-1})-\sqrt{x}(\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x-1}})}{(\sqrt{x}-\sqrt{x-1})^{2}}$Books answers:

$f'(x)=\frac{5\cdot (x+1)^{9}\cdot (x-1)}{x^{6}}$

$g'(x)=1+\frac{2x-1}{2\sqrt{x^{2}-x}}$Thank you !

For the first function:

$$\displaystyle f(x)=\left(x^{\Large\frac{1}{2}}+x^{-\Large\frac{1}{2}}\right)^{10}$$

Using the power and chain rules, I get:

$$\displaystyle f'(x)=10\left(x^{\Large\frac{1}{2}}+x^{-\Large\frac{1}{2}}\right)^{9}\left(\frac{1}{2}x^{-\Large\frac{1}{2}}-\frac{1}{2}x^{-\Large\frac{3}{2}}\right)$$

Factor:

$$\displaystyle f'(x)=5x^{-\Large\frac{3}{2}}(x-1)\left(x^{\Large\frac{1}{2}}+x^{-\Large\frac{1}{2}}\right)^{9}$$

Now, we can write:

$$\displaystyle f'(x)=5x^{-\Large\frac{3}{2}}(x-1)\left(\frac{x+1}{\sqrt{x}}\right)^{9}$$

$$\displaystyle f'(x)=5x^{-6}(x-1)(x+1)^9=\frac{5(x-1)(x+1)^9}{x^6}$$

For the second function, let's rationalize the denominator before differentiating:

$$\displaystyle g(x)=\frac{\sqrt{x}}{\sqrt{x}-\sqrt{x-1}}\cdot\frac{\sqrt{x}+\sqrt{x-1}}{\sqrt{x}+\sqrt{x-1}}=x+\sqrt{x^2-x}$$

So, then:

$$\displaystyle g'(x)=1+\frac{2x-1}{2\sqrt{x^2-x}}$$

Yankel said:
$f(x)=(\sqrt{x}+\frac{1}{\sqrt{x}})^{10}$

For the first function it also makes sense to rationalize first:
$f(x)=\left(\frac{x}{\sqrt{x}}+\frac{1}{\sqrt{x}}\right)^{10} =\left(\frac{x+1}{\sqrt{x}}\right)^{10} =\frac{(x+1)^{10}}{x^{5}}$
making it a bit easier to find the derivative.

We can do the same rationalization afterwards giving the same result.

## 1. What is differentiation with square roots?

Differentiation with square roots is the process of finding the rate of change of a function that contains a square root term. It involves using the rules of differentiation, such as the power rule, chain rule, and quotient rule, to find the derivative of the function.

## 2. Why is differentiation with square roots useful?

Differentiation with square roots is useful in many real-world applications, such as in physics and engineering, where quantities often involve square roots. It allows us to find the instantaneous rate of change of a function and can help us optimize functions to find maximum or minimum values.

## 3. How do you differentiate a square root function?

To differentiate a square root function, we can use the power rule for rational exponents. We first rewrite the function using exponent notation, then apply the power rule to find the derivative. We may also need to use the chain rule if the square root is nested within another function.

## 4. Can you provide an example of differentiation with square roots?

For example, let's find the derivative of the function f(x) = √(x^2 + 3x). Using the power rule, we can rewrite the function as f(x) = (x^2 + 3x)^1/2 and then apply the rule to get the derivative f'(x) = 1/2(x^2 + 3x)^-1/2 * (2x + 3) = (2x + 3)/2√(x^2 + 3x).

## 5. What are some common mistakes to avoid when differentiating with square roots?

Some common mistakes to avoid when differentiating with square roots include not using the correct rules of differentiation, forgetting to apply the chain rule when necessary, and not simplifying the final answer. It is also important to be careful with negative signs and to check for potential domain restrictions when simplifying the derivative.

Replies
2
Views
2K
Replies
3
Views
2K
Replies
6
Views
2K
Replies
4
Views
1K
Replies
12
Views
1K
Replies
5
Views
2K
Replies
3
Views
1K
Replies
12
Views
2K
Replies
2
Views
1K
Replies
3
Views
4K