Help: Implicit differentiation with initial values

In summary, by implicitly differentiating the given equation and simplifying, it can be shown that dy/dx = 16 when x = 1. This can be further verified by substituting y = π, where π is the only value that satisfies the equation.
  • #1
H2instinct
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Homework Statement


2*y + sin(y) = x^4 + 4(x)^3 + (2(Pi) - 5), show that dy/dx = 16, when x = 1.


Homework Equations





The Attempt at a Solution


So I implicitly differentiated it to be dy/dx(2 + cos(y)) = 4(x)^3 + 12(x)^2, and I end up with
dy/dx = 16 / (2 + cos (y)) which means that y must be equal to Pi for this to be true, but I do not think this is where I was supposed to go with this problem. Is there a way to factor or substitute out Cos(y) to show that dy/dx = 16, when x = 1?
 
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  • #2
[tex] \frac{dy}{dx} = \frac{4x^3+12x^2}{2+\cos(y)} [/tex]

[tex] 2y+\sin(y) = 2\pi [/tex]

[tex] y=\pi \iff 2\pi +\sin(\pi) = 2\pi [/tex]

[tex] \frac{dy}{dx} = \frac{16}{2+\cos(\pi)}=16 [/tex]
 
  • #3
This is exactly what I had before. Is there any way of making this proof without first implying that y must = Pi?

Edit: I see where the proof comes from. I got it correct without realizing it. Thanks for the help
 
Last edited:

1. What is implicit differentiation?

Implicit differentiation is a method used in calculus to find the derivative of an equation that is not in the form of y = f(x). It involves treating both the dependent and independent variables as functions and using the chain rule to find the derivative.

2. How is implicit differentiation different from explicit differentiation?

In explicit differentiation, the dependent variable y is explicitly written as a function of x, making it easier to find the derivative. In implicit differentiation, the dependent variable y is not explicitly written as a function of x, making it more challenging to find the derivative.

3. What are initial values in implicit differentiation?

Initial values refer to the values of x and y at a specific point on the function. These values are used to find the derivative and can be given in the problem or calculated using the given information.

4. How do I use initial values in implicit differentiation?

To use initial values in implicit differentiation, substitute the given values of x and y into the derivative equation and solve for the derivative. These values help to find the slope of the tangent line at a specific point on the function.

5. Can I use implicit differentiation to solve real-world problems?

Yes, implicit differentiation can be used to solve real-world problems that involve rates of change, such as finding the maximum or minimum values of a function or determining the velocity and acceleration of an object at a specific point. It is a valuable tool in many fields, including physics, engineering, and economics.

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