Implicit differentiation - Help

In summary, Homework Equations: -The Attempt at a Solution: cos y=x-12d/dx cos y = d/dx(12-x2)dy/dx=-sin y = -2xdy/dx=2x*cosec(x)The first differential is easy, it gets messy from here.d2y/dx2=d/dx*dy/dx=d/dx 2x (sin y)-1 =2/(sin y) +2x d/dx (sin y)-1 -product=2/(sin y) +2x [dy/dx -d
  • #1
pat666
709
0

Homework Statement


Find d2y/dx2 for cos y + x2 =12
The first derivative I think is correct but the second I am unsure of, this is Q5 of 10 in an assignment and every other question has been far cleaner and easier making me think that I've missed something.
We don't have to use implicit differentiation, I thought that it would be easiest though.
thanks for any future help.


Homework Equations





The Attempt at a Solution


cos y=x-12
d/dx cos y = d/dx(12-x2)
dy/dx=-sin y = -2x
dy/dx=2x*cosec(x)
The first differential is easy, it gets messy from here.
d2y/dx2=d/dx*dy/dx
=d/dx 2x (sin y)-1
=2/(sin y) +2x d/dx (sin y)-1 -product
=2/(sin y) +2x [dy/dx -d/dy(sin y)-1]
=2/sin y = (-4x2 cos y)/(sin y)2
=2 cosec y + -4x2 cot y cosec2 y
 
Physics news on Phys.org
  • #2
dy/dx=-sin y = -2x

This line confuses me... Anyways let's give this a whirl!

[tex]\frac{d(cos(y) + x^{2}=12)}{dx}[/tex]

[tex] 2x - y'sin(y) = 0 [/tex]

EDIT: Whoops I made a mistake after this point, I'll try to post the correct solution in my next post.:blushing:
 
Last edited:
  • #3
thanks a lot jegues. Sorry if this is a bit rude but how sure are you that this is right? I only ask because the assignment is 20% of my final grade (ridiculous I know).
 
  • #4
Wait on that chain rule step you have to use the product rule again: y' sin y isn't a function of sin y.

Anyway, OP, your answer is right, assuming you're allowed to express your answer implicitly. You can also set y = arccos(12-x2) and just differentiate directly, which after some mildly ugly algebra, results in an analytic expression in terms of x.
 
  • #5
hgfalling said:
Anyway, OP, your answer is right, assuming you're allowed to express your answer implicitly. You can also set y = arccos(12-x2) and just differentiate directly, which after some mildly ugly algebra, results in an analytic expression in terms of x.
That was my first attempt but I have up when it got to ugly, also what is OP??
 
  • #6
OP = "original poster", a common phrase on forums I've inhabited before, but not that common here now that I think about it.
 
  • #7
OP = Original Poster

And personally, I would take the derivatives implicitly. It's not that hard, and jegues was even nice enough to give you the first derivative.

However, just to let you know, posting the complete solution is against the PF rules. We want our friends to understand where the answer comes from, and such. I'm sure you understand.

From his first derivative, though, it should be easy to get the second.
 
  • #8
OP, your answer is right

Sorry about the confusion, after going through my steps again I realized I had some mistakes. You are indeed right.
 
  • #9
alright, now I am confused am I right or wrong?
 
  • #10
thanks everyone - I am now assuming my OP was right.
 
  • #11
alright, now I am confused am I right or wrong?

You are correct.

2 cosec y -4x2 cot y cosec2 y is the right answer.

Again, sorry for any confusion I may have caused.
 

1. What is implicit differentiation?

Implicit differentiation is a technique used to find the derivative of a function that is not explicitly written in terms of its independent variable. This method involves differentiating both sides of an equation with respect to the independent variable, treating the dependent variable as a function of the independent variable.

2. When is implicit differentiation used?

Implicit differentiation is used when the dependent variable cannot be easily isolated on one side of the equation. This often occurs when the equation is non-linear or contains multiple variables.

3. How does implicit differentiation differ from explicit differentiation?

Explicit differentiation is used when the dependent variable is explicitly written in terms of the independent variable, making it easier to isolate and differentiate. Implicit differentiation, on the other hand, is used when the dependent variable cannot be easily isolated and requires the use of the chain rule to differentiate.

4. What is the chain rule in implicit differentiation?

The chain rule is a calculus rule used to find the derivative of a composite function. In implicit differentiation, the chain rule is used to differentiate the dependent variable with respect to the independent variable.

5. Can implicit differentiation be used to find higher-order derivatives?

Yes, implicit differentiation can be used to find higher-order derivatives by repeating the process multiple times. However, the calculations can become more complex and time-consuming as the order of the derivative increases.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
718
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
895
  • Calculus and Beyond Homework Help
Replies
24
Views
1K
  • Calculus and Beyond Homework Help
Replies
25
Views
312
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
819
  • Calculus and Beyond Homework Help
Replies
1
Views
736
  • Calculus and Beyond Homework Help
Replies
6
Views
838
  • Calculus and Beyond Homework Help
Replies
1
Views
483
Back
Top