Finding The Nth Derivative Of 1/[(3x-2)(x-3)^2] - A Calculus Deep Dive

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Hey guys! Today, we're diving into a fun and challenging calculus problem: finding the nth derivative of the function 1/[(3x-2)(x-3)^2]. Buckle up, because this involves a mix of partial fractions, some clever differentiation techniques, and a bit of pattern recognition. Trust me, it's gonna be worth it! Let's break this down step-by-step so it’s super clear.

1. Setting the Stage: Partial Fraction Decomposition

Before we even think about derivatives, we need to simplify our function. The key here is partial fraction decomposition. This technique allows us to break down complex rational functions into simpler fractions that are much easier to work with. So, let's start with our function:

f(x) = 1 / [(3x - 2)(x - 3)^2]

Our goal is to express f(x) in the form:

f(x) = A / (3x - 2) + B / (x - 3) + C / (x - 3)^2

where A, B, and C are constants that we need to find. To do this, we'll multiply both sides of the equation by the original denominator, (3x - 2)(x - 3)^2. This gives us:

1 = A(x - 3)^2 + B(3x - 2)(x - 3) + C(3x - 2)

Now, we can solve for A, B, and C by strategically choosing values for x. Let's start with x = 3. This eliminates the terms with A and B, making it easy to solve for C:

1 = A(0) + B(0) + C(3(3) - 2)
1 = 7C
C = 1/7

Next, let's use x = 2/3. This eliminates the terms with B and C, allowing us to find A:

1 = A((2/3) - 3)^2 + B(0) + C(0)
1 = A(-7/3)^2
1 = A(49/9)
A = 9/49

To find B, we can pick any other value for x. A simple choice is x = 0. Plugging this in, along with the values we found for A and C, gives us:

1 = A(-3)^2 + B(-2)(-3) + C(-2)
1 = 9A + 6B - 2C
1 = 9(9/49) + 6B - 2(1/7)
1 = 81/49 + 6B - 2/7

Solving for B:

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:

f(x) = (9/49) / (3x - 2) - (3/49) / (x - 3) + (1/7) / (x - 3)^2

Now that we've decomposed the function, it's time to tackle those derivatives!

2. Finding the First Few Derivatives: Spotting the Pattern

Okay, guys, let's roll up our sleeves and get our hands dirty with some actual differentiation. We're going to find the first few derivatives of each term in our decomposed function and see if we can spot a pattern. This is crucial for figuring out a general formula for the nth derivative.

Let's start with the first term:

u(x) = (9/49) / (3x - 2) = (9/49)(3x - 2)^{-1}
  • First Derivative (u'(x)):
u'(x) = (9/49)(-1)(3x - 2)^{-2}(3) = -(27/49)(3x - 2)^{-2}
  • Second Derivative (u''(x)):
u''(x) = -(27/49)(-2)(3x - 2)^{-3}(3) = (162/49)(3x - 2)^{-3}
  • Third Derivative (u'''(x)):
u'''(x) = (162/49)(-3)(3x - 2)^{-4}(3) = -(1458/49)(3x - 2)^{-4}

Notice a pattern? The coefficient is changing, and we're getting alternating signs. Also, the power of (3x - 2) is increasing by one in the negative direction each time. The constant 3 keeps popping up from the chain rule, which is super important.

Now, let's look at the second term:

v(x) = -(3/49) / (x - 3) = -(3/49)(x - 3)^{-1}
  • First Derivative (v'(x)):
v'(x) = -(3/49)(-1)(x - 3)^{-2} = (3/49)(x - 3)^{-2}
  • Second Derivative (v''(x)):
v''(x) = (3/49)(-2)(x - 3)^{-3} = -(6/49)(x - 3)^{-3}
  • Third Derivative (v'''(x)):
v'''(x) = -(6/49)(-3)(x - 3)^{-4} = (18/49)(x - 3)^{-4}

Again, we see alternating signs and increasing powers of (x - 3) in the negative direction. This one's a bit simpler because there's no constant multiplying x inside the parentheses.

Finally, let's tackle the third term:

w(x) = (1/7) / (x - 3)^2 = (1/7)(x - 3)^{-2}
  • First Derivative (w'(x)):
w'(x) = (1/7)(-2)(x - 3)^{-3} = -(2/7)(x - 3)^{-3}
  • Second Derivative (w''(x)):
w''(x) = -(2/7)(-3)(x - 3)^{-4} = (6/7)(x - 3)^{-4}
  • Third Derivative (w'''(x)):
w'''(x) = (6/7)(-4)(x - 3)^{-5} = -(24/7)(x - 3)^{-5}

Same story here! Alternating signs and increasing negative powers. We're definitely seeing some patterns emerge, which is exactly what we want. Now comes the fun part: generalizing these patterns into a formula for the nth derivative.

3. Generalizing the Pattern: The nth Derivative Formula

Alright, guys, this is where the magic happens! We've computed the first few derivatives of each term, and now we're going to piece together the puzzle to find a general formula for the nth derivative. This involves a bit of inductive reasoning and careful observation.

Let's revisit our decomposed function:

f(x) = (9/49)(3x - 2)^{-1} - (3/49)(x - 3)^{-1} + (1/7)(x - 3)^{-2}

We'll find the nth derivative for each term separately and then add them up.

3.1 nth Derivative of (9/49)(3x - 2)^-1

From our earlier calculations, we saw a pattern for u(x) = (9/49)(3x - 2)^-1. Let's generalize it. The nth derivative, u^(n)(x), looks like this:

u^(n)(x) = (9/49)(-1)^n (n!) (3)^{n} (3x - 2)^{-(n+1)}

Let's break this down:

  • (9/49): This is the original coefficient, which stays constant.
  • (-1)^n: This accounts for the alternating signs. When n is even, the sign is positive; when n is odd, it's negative.
  • n!: This is n factorial (n! = n × (n-1) × (n-2) × ... × 1), which comes from the repeated application of the power rule.
  • 3^n: This comes from the chain rule. Each time we differentiate, we multiply by the derivative of (3x - 2), which is 3.
  • (3x - 2)^-(n+1): The power increases by one (in the negative direction) with each differentiation.

So, we've got a solid formula for the nth derivative of the first term. Awesome!

3.2 nth Derivative of -(3/49)(x - 3)^-1

Now, let's tackle the second term, v(x) = -(3/49)(x - 3)^-1. The nth derivative, v^(n)(x), will have a similar form, but without the 3^n factor (because the derivative of (x - 3) is just 1):

v^(n)(x) = -(3/49)(-1)^n (n!) (x - 3)^{-(n+1)}

Again, let's break it down:

  • -(3/49): The original coefficient.
  • (-1)^n: Alternating signs.
  • n!: From the power rule.
  • (x - 3)^-(n+1): The power increases by one (in the negative direction).

Notice the similarities and differences compared to the first term. The key is understanding where each part of the formula comes from.

3.3 nth Derivative of (1/7)(x - 3)^-2

Finally, let's find the nth derivative of the third term, w(x) = (1/7)(x - 3)^-2. This one's a little different because we start with a power of -2. The nth derivative, w^(n)(x), looks like this:

w^(n)(x) = (1/7)(-1)^n (n + 1)! (x - 3)^{-(n+2)}

Let's break it down:

  • (1/7): The original coefficient.
  • (-1)^n: Alternating signs.
  • (n + 1)!: This is (n+1) factorial, which is (n+1) × n × (n-1) × ... × 1. It's different from the previous terms because we started with a power of -2.
  • (x - 3)^-(n+2): The power increases by one (in the negative direction) with each differentiation, starting from -2.

See how the factorial term and the power of (x - 3) are adjusted because of the initial power of -2? This is crucial for getting the formula right.

4. Putting It All Together: The Final Formula

Okay, guys, drumroll please! We've found the nth derivative of each term in our decomposed function. Now, we just need to add them up to get the nth derivative of the original function, f(x). Here it is:

f^(n)(x) = (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)}

This might look a bit intimidating, but remember, we built it up piece by piece. Each term comes directly from the patterns we observed in the individual derivatives.

We can simplify this a bit by factoring out some common terms:

f^(n)(x) = (-1)^n n! [ (9/49) (3)^{n} (3x - 2)^{-(n+1)} - (3/49) (x - 3)^{-(n+1)} + (1/7) (n + 1) (x - 3)^{-(n+2)} ]

This is a more compact form, but both versions are correct.

5. Wrapping Up: Why This Matters

So, there you have it! We've successfully found the nth derivative of 1/[(3x-2)(x-3)^2]. This wasn't a walk in the park, but we tackled it using partial fraction decomposition, careful differentiation, pattern recognition, and a bit of algebraic manipulation. Pat yourselves on the back, guys!

But why does this matter? Finding higher-order derivatives is super important in many areas of math and physics. For example:

  • Taylor Series: Higher derivatives are crucial for constructing Taylor series, which are used to approximate functions.
  • Differential Equations: Many differential equations involve higher-order derivatives, and solving them can model all sorts of real-world phenomena.
  • Physics: Higher derivatives represent rates of change of rates of change (like jerk, the rate of change of acceleration), which are important in mechanics.

Understanding how to find these derivatives, especially for complex functions, is a valuable skill. And the process we used here – breaking down a problem into smaller parts, finding patterns, and generalizing – is a powerful problem-solving strategy that you can apply in many different contexts.

So, keep practicing, keep exploring, and keep those calculus muscles strong! You guys got this!