Another implicit differentiation question

In summary, the conversation discusses solving a problem involving y = 8x + 5cos(xy) + 7 and finding its derivative, dy/dx. The final solution is dy/dx = [8 - (5y) (sin xy)]/[(5x) (sin xy) + 1]. The speaker also mentions a more efficient way to solve the problem without writing out all the in-between steps.
  • #1
donjt81
71
0
ok so i did this problem but i wasnt sure if this is correct.

y = 8x + 5cos(xy) + 7

dy/dx = 8 + 5(-sin xy)(x dy/dx + y)
dy/dx = 8 - (5x) (sin xy) dy/dx - (5y) (sin xy)
(5x) (sin xy) dy/dx + dy/dx = 8 - (5y) (sin xy)
[(5x) (sin xy) + 1] dy/dx = 8 - (5y) (sin xy)
dy/dx = [8 - (5y) (sin xy)]/[(5x) (sin xy) + 1]

does that look right?
 
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  • #2
Looks fine. I don't see any errors.

I wish I could show you via this forum how to "think" to go from the first line to the last line without any of the in-between steps.

Maybe you can figure it out from this: you know that you're going to get all the dy/dx terms on one side and then factor out dy/dx later. So, the numerator simply becomes all the non dy/dx terms, and the denominator becomes what's left. Minor attention to signs, and you save a LOT of writing.
 

1. What is implicit differentiation?

Implicit differentiation is a method used in calculus to find the derivative of a function that is not explicitly defined in terms of a dependent variable. In other words, it is used to find the derivative of a function where both the dependent and independent variables are present in the equation.

2. How is implicit differentiation different from explicit differentiation?

Explicit differentiation is used to find the derivative of a function that is explicitly defined in terms of a dependent variable. This means that the function is written in the form of y = f(x). On the other hand, implicit differentiation is used when the function is not explicitly defined in terms of the dependent variable, but rather in terms of both the dependent and independent variables.

3. When is implicit differentiation used?

Implicit differentiation is used when it is not possible to solve for the dependent variable in a given equation. This can occur when the equation is too complex or when the dependent variable is not isolated on one side of the equation.

4. What are the steps for implicit differentiation?

The steps for implicit differentiation are as follows:
1. Differentiate both sides of the equation with respect to the independent variable.
2. Use the chain rule for any terms that involve the dependent variable.
3. Collect all terms with the derivative of the dependent variable on one side of the equation.
4. Solve for the derivative of the dependent variable.

5. What are some real-life applications of implicit differentiation?

Implicit differentiation has many real-life applications, including:
1. Calculating the velocity and acceleration of moving objects
2. Finding the slope of curves that are not explicitly defined
3. Solving problems in physics, engineering, and economics that involve rates of change
4. Analyzing the behavior of complex systems, such as population growth or chemical reactions.

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