Finding the derivative of a function with a radical

In summary: I went wrong, I was able to solve it from there. In summary, the conversation was about finding the derivative of a given equation using the product rule and chain rule, and the individual getting stuck at simplifying their solution. They were able to receive help and successfully solve the problem.
  • #1
waffletree
4
0

Homework Statement


e1bbd57bb281ee733177266492e355.gif

Find the differential

Homework Equations


Chain rule : dy/du=dy/du*du/dx
Product rule: f(x)g'(x) + g(x)f'(x)

The Attempt at a Solution


I have tried to move the radical to the top of the equation by making it into an exponent (x^2+1)^-1/2. I then used the product rule and the chain ruleIand i got lost somewhere simplifying it beyond that.
 
Physics news on Phys.org
  • #2
Hello Waffle, :welcome:

PF culture pushes me to ask this :smile:

Where were you when you got stuck ? In other words: show us how far you got and why you think you were lost at that point ...
 
  • #3
waffletree said:

Homework Statement


e1bbd57bb281ee733177266492e355.gif

Find the differential

Homework Equations


Chain rule : dy/du=dy/du*du/dx
Product rule: f(x)g'(x) + g(x)f'(x)

The Attempt at a Solution


I have tried to move the radical to the top of the equation by making it into an exponent (x^2+1)^-1/2. I then used the product rule and the chain ruleIand i got lost somewhere simplifying it beyond that.

Show us what you got. Maybe your final,answer is correct, and maybe it isn't; we cannot tell if you will not show it to us.
 
  • #4
waffletree said:

Homework Statement


e1bbd57bb281ee733177266492e355.gif

Find the differential

Homework Equations


Chain rule : dy/du=dy/du*du/dx
Product rule: f(x)g'(x) + g(x)f'(x)

The Attempt at a Solution


I have tried to move the radical to the top of the equation by making it into an exponent (x^2+1)^-1/2. I then used the product rule and the chain ruleIand i got lost somewhere simplifying it beyond that.
Please show us what you did. Your steps of using the product rule followed by the chain rule are the right approach.
Also, do you need to find the differential or the derivative? They are related, but not the same.
 
  • #5
Mark44 said:
Please show us what you did. Your steps of using the product rule followed by the chain rule are the right approach.
Also, do you need to find the differential or the derivative? They are related, but not the same.
Sorry I meant to put derivative,
This is where I got stuck at
6x(x^2+1)^-1/2
6x*d/dx(x^2+1)^-1/2+(x^2+1)^-1/2*6
6x*1/2(x^2+1)^-3/2+(x^2+1)^-1/2*6
 
  • #6
BvU said:
Hello Waffle, :welcome:

PF culture pushes me to ask this :smile:

Where were you when you got stuck ? In other words: show us how far you got and why you think you were lost at that point ...

Sorry I meant to put derivative,
This is where I got stuck at
6x(x^2+1)^-1/2
6x*d/dx(x^2+1)^-1/2+(x^2+1)^-1/2*6
6x*1/2(x^2+1)^-3/2+(x^2+1)^-1/2*6
 
  • #7
waffletree said:
Sorry I meant to put derivative,
This is where I got stuck at
6x(x^2+1)^-1/2
6x*d/dx(x^2+1)^-1/2+(x^2+1)^-1/2*6
6x*1/2(x^2+1)^-3/2+(x^2+1)^-1/2*6

##\frac{d}{dx} [(x^2+1)^{-1/2}] = (-1/2) (x^2+1)^{-3/2} \cdot \frac{d}{dx} (x^2+1) = -x(x^2+1)^{-3/2}##.
 
  • #8
Ray Vickson said:
##\frac{d}{dx} [(x^2+1)^{-1/2}] = (-1/2) (x^2+1)^{-3/2} \cdot \frac{d}{dx} (x^2+1) = -x(x^2+1)^{-3/2}##.
Thank you
 

1. How do you find the derivative of a function with a radical?

The first step is to rewrite the radical using exponent rules. For example, the cube root of x can be written as x^(1/3). Then, use the power rule to find the derivative, which is d/dx(x^n) = nx^(n-1). Finally, simplify the derivative as much as possible.

2. Can you use the chain rule to find the derivative of a function with a radical?

Yes, the chain rule can be used when finding the derivative of a function with a radical. Simply treat the radical as the "inner function" and apply the chain rule as you normally would.

3. Are there any special cases when finding the derivative of a function with a radical?

Yes, if the radical is a square root and there is a constant inside, you can use the power rule and treat the constant as a coefficient. If the radical is a higher root (such as a cube root), you will need to use a combination of the power rule and the chain rule.

4. Can you find the derivative of a function with a radical using the quotient rule?

Yes, the quotient rule can be used if the radical is in the denominator of a fraction. Simply apply the quotient rule as you normally would, treating the radical as the "bottom function" and the rest of the function as the "top function".

5. How does the derivative of a function with a radical compare to the derivative of a function without a radical?

The process for finding the derivative is similar, but the resulting derivative may be more complex due to the presence of the radical. It is important to simplify the derivative as much as possible to make it easier to work with.

Similar threads

The below solution seems to assume that 1/0 = 0
    • Calculus and Beyond Homework Help
Replies
7
Views
860
Calculating 2nd Total Derivative of u w.r.t. t
    • Calculus and Beyond Homework Help
Replies
28
Views
2K
Why Is the Chain Rule Not Used in Differentiating h(x) = 3f(x) + 8g(x)?
    • Calculus and Beyond Homework Help
Replies
2
Views
925
Find the length of the curve C
    • Calculus and Beyond Homework Help
Replies
1
Views
738
For this Partial Derivative -- Why are different results obtained?
    • Calculus and Beyond Homework Help
Replies
23
Views
1K
Missing some fundamental understanding- U-sub w/ radicals
    • Calculus and Beyond Homework Help
Replies
8
Views
994
Finding limits with a radical in denominator
    • Calculus and Beyond Homework Help
Replies
5
Views
1K
Find the derivative of the given function
    • Calculus and Beyond Homework Help
Replies
4
Views
843
Solve the given differential equation
    • Calculus and Beyond Homework Help
Replies
10
Views
1K
The derivative of uv wrt x using st function (homework problem)
    • Calculus and Beyond Homework Help
Replies
6
Views
740

Hot Threads

Recent Insights

Back
Top