How do you find the derivative of a function involving square roots?

In summary, you can differentiate a function like this: f(x) = sqrt( 1 - x² ) by treating x as a constant and applying the chain rule.
  • #1
the1024b
5
0
How can i find the derivative of a function like this:
f(x) = sqrt( 1 - x² )
 
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  • #2
Do you know how to write a square root with exponents?
 
  • #3
(1 - x² )^(1/2) ?
 
  • #4
That's right! Now, you just need to apply what you know about differentiating expressions like that.
 
  • #5
si will that be:
1/2((1-x²)/2)^(-1/2)

?
 
  • #6
Not quite. You have one too many "1/2"s (you don't want that "/2" inside the square root and you didn't use the chain rule.

You need to multiply by the derivative of 1-x2.
 
Last edited by a moderator:
  • #7
Hi,

I have a similar problem, I need to differentiate sqrt(x^2 + y^2) in terms of x and y.
Starting this I took the simple step (x^2 + y^2)^(1/2)...

My next step is a guess and I am lost after it...(1/2)(x^2 + y^2)(-1/2)...

Any help would be much appreciated.
 
  • #8
mathsn00b said:
Hi,

I have a similar problem, I need to differentiate sqrt(x^2 + y^2) in terms of x and y.
Starting this I took the simple step (x^2 + y^2)^(1/2)...

My next step is a guess and I am lost after it...(1/2)(x^2 + y^2)(-1/2)...

Any help would be much appreciated.

If by "in terms of x and y", you mean you want to calculate the partial derivatives, then for the partial derivative with respect to x, treat y as a constant and differentiate with respect to x as you normally would a function of one variable. For the partial derivative with respect to y, treat x as constant.
 
  • #9
thanks, would I do this by...

df/dx = 1/2(x^2 + y^2)^(-1/2).2x = x/sqrt(x^2 + y^2) and...

df/dy = 1/2(x^2 + y^2)^(-1/2).2y = y/sqrt(x^2 + y^2) ?

thanks for your help so quickly.
 
  • #10
mathsn00b said:
thanks, would I do this by...

df/dx = 1/2(x^2 + y^2)^(-1/2).2x = x/sqrt(x^2 + y^2) and...

df/dy = 1/2(x^2 + y^2)^(-1/2).2y = y/sqrt(x^2 + y^2) ?

thanks for your help so quickly.

Looks good to me.
 
  • #11
What if i have a problem similar to these however now its 4/ ^5sqrt(x^5)
 
  • #12
68Pirate said:
What if i have a problem similar to these however now its 4/ ^5sqrt(x^5)
If that is meant to be 4^(5(sqrt(x^5))), then you can easily rewrite this to equal
4^(5(x^(5/2)) And using what you know from differentiating exponentials and chain rule, you should be able to get the rest.
 

1. What is a derivative of a square root?

A derivative of a square root is the rate of change of a square root function at a specific point. It represents the slope of the tangent line to the square root function at that point.

2. How do you find the derivative of a square root?

To find the derivative of a square root, you can use the power rule and chain rule. First, rewrite the square root as an exponent of 1/2. Then, apply the power rule and chain rule to find the derivative.

3. What is the derivative of the square root of x?

The derivative of the square root of x is 1/2 times the reciprocal of the square root of x. In other words, it is 1/2 divided by the square root of x.

4. Can you provide an example of finding the derivative of a square root?

Sure, let's find the derivative of f(x) = √x at x = 4. First, rewrite the square root as x^(1/2). Then, use the power rule and chain rule to find the derivative: f'(x) = 1/2 * x^(-1/2) * 1 = 1/2√x. Substituting x = 4, we get f'(4) = 1/2√4 = 1/4.

5. Is the derivative of a square root always positive?

No, the derivative of a square root can be positive, negative, or zero, depending on the value of x. For example, at x = 4, the derivative of √x is positive (1/4), but at x = 0, the derivative is undefined. It is important to consider the domain of the function when determining the sign of the derivative.

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