Complicated implicit differentiation

In summary, the problem involves finding the derivative, dy/dx, of an implicit function defined by the equation (sqrt x) + (sqrt y) = 11, and using the given information of f(9)=64 to solve for dy/dx. The correct solution is -8/3.
  • #1
cal.queen92
43
0

Homework Statement



If (sqrt x) + (sqrt y) = 11 and f(9)=64 ---> find f '(9) by implicit differentiation



The Attempt at a Solution



I keep getting lost in my work here...

first, taking derivative of both sides:

d/dx ((sqrt x) + (sqrt y)) = d/dx (11)

obtaining: (1/2)(x)^(-1/2) + (1/2)(y)^(-1/2) * dy/dx = 0

Now, I want to keep y positive so:

(1/2)(y)^(-1/2) * dy/dx = -(1/2)(x)^(-1/2)

So if I solve for dy/dx:

dy/dx = (-1/(sqrt x)) / (1/(sqrt y)) which means: dy/dx = (-1/(sqrt x) * ((sqrt y)/1)

giving: dy/dx = -(sqrt y)/ (sqrt x) as the derivative.

However, I don't know how to use the other information provided! I am very stuck... If anyone has any ideas, thanks!
 
Physics news on Phys.org
  • #2
cal.queen92 said:

Homework Statement



If (sqrt x) + (sqrt y) = 11 and f(9)=64 ---> find f '(9) by implicit differentiation



The Attempt at a Solution



I keep getting lost in my work here...

first, taking derivative of both sides:

d/dx ((sqrt x) + (sqrt y)) = d/dx (11)

obtaining: (1/2)(x)^(-1/2) + (1/2)(y)^(-1/2) * dy/dx = 0

Now, I want to keep y positive so:

(1/2)(y)^(-1/2) * dy/dx = -(1/2)(x)^(-1/2)

So if I solve for dy/dx:

dy/dx = (-1/(sqrt x)) / (1/(sqrt y)) which means: dy/dx = (-1/(sqrt x) * ((sqrt y)/1)

giving: dy/dx = -(sqrt y)/ (sqrt x) as the derivative.

However, I don't know how to use the other information provided! I am very stuck... If anyone has any ideas, thanks!

Your derivative is correct. The information you didn't use is f(9)=64. Here f is the implicit function defined by the equation x1/2 + y1/2 = 11.

f(9) = 64 means that the point (9, 64) is on the graph of the implicit function. What you're supposed to do to finish this problem is to find dy/dx when x = 9 and y = 64.
 
  • #3
I see! That's great! so then that gives an answer of -8/3. Perfect! Thank you.
 

Related to Complicated implicit differentiation

1. What is complicated implicit differentiation?

Complicated implicit differentiation is a mathematical technique used to find the derivative of a function that cannot be easily expressed in terms of one variable.

2. When is complicated implicit differentiation used?

It is used when the function is written in an implicit form, where both the dependent and independent variables are present in the equation.

3. How is complicated implicit differentiation different from regular differentiation?

In regular differentiation, the dependent variable is explicitly expressed in terms of the independent variable. In complicated implicit differentiation, the dependent variable is not explicitly expressed and must be solved for using the chain rule and other techniques.

4. What are the steps for performing complicated implicit differentiation?

The steps for performing complicated implicit differentiation include identifying the dependent and independent variables, taking the derivative of both sides of the equation, using the chain rule and other derivative rules to solve for the dependent variable, and simplifying the resulting equation.

5. What are some common mistakes to avoid when using complicated implicit differentiation?

Some common mistakes to avoid when using complicated implicit differentiation include forgetting to use the chain rule, not properly identifying the dependent and independent variables, and making algebraic errors during the simplification process.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
675
  • Calculus and Beyond Homework Help
Replies
2
Views
879
  • Calculus and Beyond Homework Help
Replies
6
Views
919
  • Calculus and Beyond Homework Help
Replies
14
Views
639
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
21
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
618
  • Calculus and Beyond Homework Help
Replies
4
Views
847
  • Calculus and Beyond Homework Help
Replies
1
Views
905
  • Calculus and Beyond Homework Help
Replies
5
Views
868
Back
Top