How Does the Derivative Step Work in Kreyszig's Solution for Y'=(y+9x)^2?

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In summary, the conversation discusses a problem from Kreyszig's 9E of advanced engineering mathematics. The problem involves finding a closed form solution, which turns out to be y=3tan(3x+c)-9x. The person has trouble with a specific step where the derivative is taken with respect to a variable. They realize their mistake and understand that y is a function of x, so the derivative should be taken as dy/dx.
  • #1
TRAyres
5
1
This problem is out of Kreyszig's 9E of advanced engineering mathematics.

I don't understand their closed form solution.
Their solution ends up being
y=3tan(3x+c)-9x.

But when they were solving v=y+9x for y,
then taking the derivative, they get:
y'=v'-9, then set that =v^2

--Thats the step I have issues with. They are taking the derivative,
but with respect to what? If it is some dummy variable,
it would be y'=v'-9x' (chain rule?). If it is y with respect to v,
the derivative of y=v-9x would just be y'=v'. Unless x' with respect to
a dummy variable is = 1, of course.

Can someone please enlighten me?
 
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  • #2
The problem simply takes dv/dx = d/dx(y+9x) = dy/dx+9 or y'+9 where ' denotes differentiation with respect to x in this case.
 
  • #3
...oh man I was treating y as a constant and not as a variable. y is a function of x, so d/dx(y)=dy/dx...

: smacks forehead : thanks.
 

FAQ: How Does the Derivative Step Work in Kreyszig's Solution for Y'=(y+9x)^2?

What is the equation "Y'=(y+9x)^2" used for?

The equation "Y'=(y+9x)^2" is used to find the derivative of a function with two variables, y and x. It is also known as the chain rule, where the derivative of the outer function is multiplied by the derivative of the inner function.

How do you solve for Y'=(y+9x)^2?

To solve for Y'=(y+9x)^2, you need to use the power rule of differentiation. This means that you multiply the exponent by the coefficient, and then subtract 1 from the original exponent. In this case, the derivative would be Y'=2(y+9x)*9, which simplifies to Y'=18(y+9x).

What is the significance of (y+9x=v) in the equation?

The equation (y+9x=v) is known as the implicit equation, where the variable "v" represents the dependent variable y. This means that the equation cannot be easily solved for y, and instead the derivative is used to find the rate of change of y with respect to x.

Can the equation "Y'=(y+9x)^2" be applied to any function?

Yes, the equation "Y'=(y+9x)^2" can be applied to any function with two variables, as long as the function can be written in the form (y+9x)=v. This allows the chain rule to be applied and the derivative to be found.

How is the equation "Y'=(y+9x)^2" related to the concept of slope?

The equation "Y'=(y+9x)^2" is related to the concept of slope as it represents the instantaneous rate of change of the function at a specific point. This is similar to the slope of a line, which represents the rate of change of y with respect to x over a given interval. However, the derivative gives the exact rate of change at a specific point on the function.

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