The derivative of r^2 dC/dr is, by the product rule, [r^2 d(dC/dr)/dr]+ [d(r^2)/dr] dC/dr= r^2 d^2C/dr+ 2r dC/dr. Multiplying that by 1/r^2 gives d^2C/dr^2+ 2/r dC/dr so the equation is the same as D(d^2C/dr^2+ 2/r dC/dr)- kC= 0.juice34 said:My question is how do they transform equation A. into B.. I know they are using the product rule but don't know what is going on.
EQ A.)D(1/r^2)d/dr(r^2(dC/dr))-kC=0
now how do they get Eq B.
EQ B.) (d^2C/dr^2)+(2/r)(dC/dr)-(kC)/D=0
The product rule is a fundamental rule in calculus that is used to find the derivative of a product of two functions. In the context of transforming equations, the product rule can be used to manipulate equations by breaking them down into smaller, more manageable parts.
Understanding the process of using product rule to transform equations is important because it allows us to manipulate equations in order to solve for a specific variable or simplify the equation. This is particularly useful in more complex equations where traditional algebraic methods may be difficult or impossible to use.
Yes, for example, if we have the equation y = (x^2 + 3x)(2x + 5), we can use the product rule to expand the equation and simplify it to y = 2x^3 + 13x^2 + 15x.
While the product rule is a powerful tool for transforming equations, it does have its limitations. It can only be used when dealing with products of two functions, and it may not always work for more complex equations with multiple variables.
The best way to practice and improve your skills in using product rule to transform equations is to solve a variety of practice problems. You can find many resources online, such as textbooks, worksheets, and videos, that provide examples and practice problems for you to work on. It is also helpful to seek guidance from a teacher or tutor if you are struggling with the concept.