Finding The Nth Derivative Of 1/[(3x-2)(x-3)^2] A Step-by-Step Guide
Hey guys! Ever stumbled upon a function that looks like a mathematical monster, especially when you need to find its nth derivative? Well, today we’re going to wrestle with one of those beasts: 1/[(3x-2)(x-3)^2]. Buckle up, because this journey involves partial fractions, a bit of calculus magic, and a dash of pattern recognition. We will go through the process step by step so that you can easily understand it.
Breaking Down the Beast: Partial Fraction Decomposition
Okay, so the first thing we need to do when faced with a complex rational function like this is to simplify it. Think of it as chopping a giant problem into bite-sized pieces. The key technique here is partial fraction decomposition. This method allows us to rewrite our fraction as a sum of simpler fractions, which are much easier to differentiate.
So, we start by assuming that our function can be expressed in the following form:
1/[(3x-2)(x-3)^2] = A/(3x-2) + B/(x-3) + C/(x-3)^2
Where A, B, and C are constants that we need to determine. The goal is to find these constants so that the equation holds true. To do this, we'll clear the denominators by multiplying both sides of the equation by (3x-2)(x-3)^2. This gives us:
1 = A(x-3)^2 + B(3x-2)(x-3) + C(3x-2)
Now, we have a polynomial equation. There are a couple of ways to solve for A, B, and C. One common method is to substitute specific values of x that will eliminate some of the terms. For instance, if we let x = 3, the terms with (x-3) will become zero, leaving us with an equation involving only C. Similarly, if we let x = 2/3, the terms with (3x-2) will vanish, allowing us to solve for A. Another approach is to expand the right side, collect like terms, and equate the coefficients of the corresponding powers of x on both sides. This will give us a system of linear equations that we can solve for A, B, and C.
Let’s go through the substitution method first. Setting x = 3, we get:
1 = A(3-3)^2 + B(3(3)-2)(3-3) + C(3(3)-2)
1 = 0 + 0 + 7C
C = 1/7
Next, let’s set x = 2/3:
1 = A(2/3 - 3)^2 + B(3(2/3)-2)(2/3-3) + C(3(2/3)-2)
1 = A(-7/3)^2 + 0 + 0
1 = A(49/9)
A = 9/49
Now, we need to find B. Since we've used the convenient roots, let’s use another simple value for x, like x = 0. Substituting x = 0, A = 9/49, and C = 1/7 into our equation, we get:
1 = (9/49)(0-3)^2 + B(3(0)-2)(0-3) + (1/7)(3(0)-2)
1 = (9/49)(9) + B(-2)(-3) + (1/7)(-2)
1 = 81/49 + 6B - 2/7
1 = 81/49 + 6B - 14/49
1 = 67/49 + 6B
6B = 1 - 67/49
6B = -18/49
B = -3/49
So, we’ve found A = 9/49, B = -3/49, and C = 1/7. This means we can rewrite our original function as:
1/[(3x-2)(x-3)^2] = (9/49)/(3x-2) - (3/49)/(x-3) + (1/7)/(x-3)^2
Now, this looks much more manageable! We’ve successfully broken down our mathematical beast into friendlier components.
The Calculus Magic: Differentiating the Pieces
Now that we’ve got our function in a simpler form, the next step is to find the nth derivative. Remember, the nth derivative of a function is just the result of differentiating it n times. Fortunately, we've broken our function into pieces that are relatively easy to differentiate. The key here is to recognize the pattern that emerges when we repeatedly differentiate each term.
Let's tackle each term separately. First, consider the term (9/49)/(3x-2). To differentiate this, it’s helpful to rewrite it as (9/49)(3x-2)^(-1). Now, we can apply the power rule and the chain rule.
The first derivative is:
(9/49)(-1)(3x-2)^(-2)(3) = (-27/49)(3x-2)^(-2)
The second derivative is:
(-27/49)(-2)(3x-2)^(-3)(3) = (162/49)(3x-2)^(-3)
The third derivative is:
(162/49)(-3)(3x-2)^(-4)(3) = (-1458/49)(3x-2)^(-4)
Notice the pattern? Each time we differentiate, we multiply by a negative number, a power of (3x-2), and a factor of 3 (from the chain rule). We also see the power of (3x-2) decreasing by 1 and the factorial-like term building up.
Now, let's generalize this. The nth derivative of (9/49)(3x-2)^(-1) is:
(9/49)(-1)^n * n! * 3^n * (3x-2)^(-n-1)
Cool, right? We’ve got a general formula for the nth derivative of the first term.
Let's move on to the second term, -(3/49)/(x-3), which we can rewrite as -(3/49)(x-3)^(-1). This is very similar to the first term, but without the extra factor of 3 from the chain rule.
The first derivative is:
-(3/49)(-1)(x-3)^(-2) = (3/49)(x-3)^(-2)
The second derivative is:
(3/49)(-2)(x-3)^(-3) = -(6/49)(x-3)^(-3)
The third derivative is:
-(6/49)(-3)(x-3)^(-4) = (18/49)(x-3)^(-4)
The pattern here is a bit simpler. The nth derivative of -(3/49)(x-3)^(-1) is:
-(3/49)(-1)^n * n! * (x-3)^(-n-1)
Awesome! One more term to go. Let’s look at the last term, (1/7)/(x-3)^2, which we can rewrite as (1/7)(x-3)^(-2).
The first derivative is:
(1/7)(-2)(x-3)^(-3) = (-2/7)(x-3)^(-3)
The second derivative is:
(-2/7)(-3)(x-3)^(-4) = (6/7)(x-3)^(-4)
The third derivative is:
(6/7)(-4)(x-3)^(-5) = (-24/7)(x-3)^(-5)
This pattern is slightly different, but we can still generalize it. Notice that instead of n!, we have a product of consecutive integers. The nth derivative of (1/7)(x-3)^(-2) is:
(1/7)(-1)^n * (n+1)! * (x-3)^(-n-2)
Putting It All Together: The Grand Finale
Alright, guys, we've done the hard work! We've broken down our function using partial fractions, and we've found the general formula for the nth derivative of each piece. Now, all that’s left is to add them up to get the nth derivative of the original function.
So, the nth derivative of 1/[(3x-2)(x-3)^2] is the sum of the nth derivatives of each term:
(9/49)(-1)^n * n! * 3^n * (3x-2)^(-n-1) - (3/49)(-1)^n * n! * (x-3)^(-n-1) + (1/7)(-1)^n * (n+1)! * (x-3)^(-n-2)
We can simplify this a bit by factoring out the (-1)^n term:
(-1)^n * [(9/49) * n! * 3^n * (3x-2)^(-n-1) - (3/49) * n! * (x-3)^(-n-1) + (1/7) * (n+1)! * (x-3)^(-n-2)]
And there you have it! This is the nth derivative of our original function. It looks a bit intimidating, but we got there by systematically breaking down the problem into smaller, manageable steps. Remember, partial fractions are your friend when dealing with rational functions, and pattern recognition is key when finding nth derivatives.
Conclusion
So, finding the nth derivative of 1/[(3x-2)(x-3)^2] might seem like a Herculean task at first, but by using partial fraction decomposition and carefully differentiating each term, we can arrive at a general formula. This process not only gives us the answer but also deepens our understanding of calculus techniques. Keep practicing, and these mathematical monsters will soon become much less scary! Remember guys, practice makes perfect!