Solving for y' Derivative: 2 sinxcosy=1

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In summary: Can you use the chain rule? i think that's what I was trying to do. I was taking the derivative of the outside then the derivative of the inside.You can apply the chain rule to g(y) (for example, g'(y) dy/dx), but you still have a product to sort through.so if you use the product rule you would get...(2sinx -siny') + (2cos
  • #1
BuBbLeS01
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Homework Statement


find the derivative of...
2 sinxcosy = 1


The Attempt at a Solution


(2 cosxcosy) * (cosy')
cos y' = -2 cosxcosy
y' = (-2 cosxcosy)/(cos)

I know that's not right but I am not sure where I am making the mistake.
 
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  • #2
How do you find the derivative of f(x)g(y)?
 
  • #3
EnumaElish said:
How do you find the derivative of f(x)g(y)?
use the product rule, f(x)g(y') + g(y)f(x')
 
  • #4
BuBbLeS01 said:
use the product rule, f(x)g(y') + g(y)f(x')
you didn't do the product rule right
 
  • #5
Can you use the chain rule? i think that's what I was trying to do. I was taking the derivative of the outside then the derivative of the inside.
 
Last edited:
  • #6
You can apply the chain rule to g(y) (for example, g'(y) dy/dx), but you still have a product to sort through.
 
  • #7
so if you use the product rule you would get...
(2sinx -siny') + (2cosxcosy) = 0
 
  • #8
Write out the formula for [f(x)g(y)]'. Then substitute in f(x) = 2 sin x, g(y) = cos y, f '(x) = ... and g'(y) = ...
 
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  • #9
f(x)g(y') + f(x')g(y)
sinxy' + 2cosxcosy
 
  • #10
okay that's not right... it should be...
2cosxcosy - 2sinxsinyy' = 0
2cosxcosy - 2y'sinxsiny = 0
y' = -2cosxcosy/-2sinxsiny
 
  • #11
BuBbLeS01 said:
f(x)g(y') + f(x')g(y)
It should be f(x)g'(y)y' + f '(x)g(y).
 
  • #12
oh ok...so is that answer right now?
2cosxcosy - 2sinxsinyy' = 0
2cosxcosy - 2y'sinxsiny = 0
y' = -2cosxcosy/-2sinxsiny
 
  • #13
Do you think it is right?

Write out f(x)g'(y)y' + f '(x)g(y).

State f(x) and g(x).

State f '(x) and g'(y).

Make the substitutions.
 
  • #14
yes I do think its right...

2sinxcosy
f(x) = 2sinx
g(x) = cosy
f'(x) = 2cosx
g'(y) = -sinyy'

2cosxcosy - 2sinxsinyy' = 0
2cosxcosy - 2y'sinxsiny = 0
y' = -2cosxcosy/-2sinxsiny
 
  • #15
I think you're right. You can cancel out the minuses.
 
  • #16
ok thank you!
 
  • #17
BuBbLeS01 said:
ok thank you!
you still have 1 more step

what is cosine/sine?
 
  • #18
BuBbLeS01 said:

Homework Statement


find the derivative of...
2 sinxcosy = 1


The Attempt at a Solution


(2 cosxcosy) * (cosy')
cos y' = -2 cosxcosy
y' = (-2 cosxcosy)/(cos)

I know that's not right but I am not sure where I am making the mistake.
[2 sinxcosx]'= 2 [(sin(x)' cos(x)+ sin(x) cos(x)']
 

FAQ: Solving for y' Derivative: 2 sinxcosy=1

What is the purpose of solving for y' derivative?

The purpose of solving for y' derivative is to find the rate of change of a function with respect to its independent variable. It helps us understand the behavior of the function and how it changes over a given interval.

How do you solve for y' derivative in a given equation?

To solve for y' derivative in a given equation, you need to use the rules of differentiation, such as the power rule, product rule, and chain rule. These rules help simplify the equation and find the derivative of each term.

Can you explain the steps involved in solving for y' derivative?

The steps involved in solving for y' derivative are as follows: first, use the rules of differentiation to simplify the equation. Next, find the derivative of each term with respect to the independent variable. Then, substitute the values of the independent variable into the derivative expression. Finally, simplify the resulting expression to get the y' derivative.

What is the significance of solving for y' derivative?

Solving for y' derivative is significant because it helps us understand the behavior of a function and how it changes over a given interval. It also allows us to find the slope of a tangent line at a specific point on the function, which is useful in many real-world applications.

Are there any common mistakes to avoid when solving for y' derivative?

Yes, there are some common mistakes to avoid when solving for y' derivative. These include forgetting to use the rules of differentiation, making errors in the algebraic simplification, and not substituting the values of the independent variable correctly. It is important to double-check your work and be careful with your calculations to avoid these mistakes.

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