Implicit Differentiation problem

In summary: Is it possible for you to explain to me how you got to each step? Sure thing! So, the first step is just the initial equation, where we take the derivative of both sides with respect to x. Then, in the second step, we use the product rule to take the derivative of y^2 and x*sin(y). In the third step, we simplify and combine like terms. In the fourth step, we factor out a dy/dx term, and then in the fifth step, we solve for it by dividing both sides by (2y+x*cos(y)). And finally, in the last step, we get the final answer by simplifying. Does that make sense?
  • #1
suxatphysix
30
0

Homework Statement



Find dy/dx by implicit differentiation when it is known that y^2 + xsiny = 4

Homework Equations





The Attempt at a Solution



2y dy/dt + xcosy dy/dt + siny = 0

2y dy/dt + xcosy dy/dt = -siny

dy/dt + dy/dt = -siny/2y/xcosy

I'm sure I'm doing it wrong so I stopped right there. What am I doing it wrong and how do I solve it?

Thanks
 
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  • #2
you're looking for dy/dx not dy/dt o.o, but assuming by dt you mean dx then in the 2nd line of your attempt you should pull out the dy/dt and the answer will be easy from there.
 
  • #3
Is this the answer?

dy/dx = -siny/2y+xcosy
 
  • #4
It would be if you would parenthesize the denominator.
 
  • #5
dy/dx = -siny/(2y+xcosy) ? What's that's do
 
  • #6
It distinguishes between -(siny/(2y))+(xcosy) and -siny/(2y+xcosy). Which are two quite different expressions. But look the same if you omit the parentheses.
 
Last edited:
  • #7
1. Homework Statement

Find dy/dx by implicit differentiation when it is known that y^2 + xsiny = 4

[tex]\frac{d}{dx}(y^2+x\sin(y)) = 0[/tex]

[tex]=2y\left(\frac{dy}{dx}\right)+\left[\frac{d}{dx}x\cdot\sin(y)+\frac{d}{dx}\sin(y)\cdot x\right]=0[/tex]

[tex]= 2y\left(\frac{dy}{dx}\right) + \left[\sin(y) + x\cdot\cos(y)\left(\frac{dy}{dx}\right)\right]= 2y\left(\frac{dy}{dx}\right) + \sin(y) + x\cos(y)\left(\frac{dy}{dx}\right) = 2y\left(\frac{dy}{dx}\right) + x\cos(y)\left(\frac{dy}{dx}\right) + \sin(y) = 0[/tex]

[tex]= \left(\frac{dy}{dx}\right)(2y+x\cos(y)) + \sin(y) = 0[/tex]

[tex]= \left(\frac{dy}{dx}\right)(2y+x\cos(y)) = -\sin(y)[/tex]

[tex]\left(\frac{dy}{dx}\right) = \frac{-\sin(y)}{(2y+x\cos(y))}[/tex]

Edit: I'm late by about 25 minutes. :biggrin:
 
  • #8
cool thanks
 

Related to Implicit Differentiation problem

What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not explicitly written in terms of one variable. It is commonly used when the dependent variable cannot be easily isolated on one side of the equation.

How is implicit differentiation different from explicit differentiation?

In explicit differentiation, the derivative is found by directly differentiating the function with respect to the independent variable. In implicit differentiation, the derivative is found by taking the derivative of both sides of an equation with respect to the independent variable, treating the dependent variable as a function of the independent variable.

When should I use implicit differentiation?

Implicit differentiation is useful when the dependent variable cannot be easily isolated on one side of the equation, or when the equation represents a curve rather than a function. It is commonly used in applications of calculus, such as finding the slope of a tangent line to a curve.

What are the steps for solving an implicit differentiation problem?

The steps for solving an implicit differentiation problem are as follows:

  1. Differentiate both sides of the equation with respect to the independent variable.
  2. Use the chain rule to differentiate any terms that involve the dependent variable.
  3. Isolate the derivative on one side of the equation.
  4. Simplify the resulting equation, if possible.
  5. Plug in any known values to find the value of the derivative at a specific point, if needed.

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

Some common mistakes to avoid when using implicit differentiation include:

  • Forgetting to use the chain rule when differentiating terms with the dependent variable.
  • Not isolating the derivative on one side of the equation before simplifying.
  • Missing or incorrect use of parentheses when differentiating multiple terms.
  • Forgetting to plug in known values at the end to find the value of the derivative at a specific point.

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