Find the derivative of f(x) = {(x - 1)/(2x² - 7x + 5) when x ≠ 1; -1/3 when x = 1} at x = 1

Q: Wait, why can't I just use the quotient rule on the original expression?

That is a classic trap! The quotient rule is designed for functions that are differentiable in an open interval around the point. At $x=1$, the original expression $\frac{x-1}{2x^2-7x+5}$ results in a $\frac{0}{0}$ indeterminate form, meaning the expression itself is undefined there. The piecewise definition provides the value at $x=1$, which is essential. The derivative at a point is defined by the limit of the difference quotient, not by differentiating the formula for $x \neq 1$.

Q: Do I need to check for continuity first?

That is a brilliant instinct! Yes, for a function to be differentiable at a point, it must first be continuous there. If the limit of $f(x)$ as $x \to 1$ did not equal $f(1)$, the function would have a jump or a hole, and the derivative would not exist. In this case, $\lim_{x \to 1} \frac{1}{2x-5} = -\frac{1}{3}$, which matches $f(1)$, confirming continuity and allowing us to proceed with confidence.

Q: What happens if I don't simplify the function first?

If you don't simplify, you will be stuck calculating a limit of a very complex fraction involving $(1+h-1)$ in the numerator and a quadratic in the denominator. While it is technically possible, it is algebraically exhausting and prone to errors. Simplifying the function to $f(x) = \frac{1}{2x-5}$ for $x \neq 1$ is the key to unlocking the problem elegantly.

Question

Find the derivative of $f(x) = \begin{cases} \frac{x-1}{2x^2-7x+5} & 	ext{when } x 
eq 1 \ -\frac{1}{3} & 	ext{when } x = 1 \end{cases}$ at $x = 1$.

Sigma Percentile · Accepted Answer

### The Beauty of the Piecewise Puzzle

Welcome, fellow traveler on the journey of calculus! Today, we are going to tackle a problem that might seem like a simple derivative exercise at first glance, but it hides a beautiful, subtle trap.

We are looking at a piecewise function defined as:

$$f(x) = \begin{cases} \frac{x-1}{2x^2-7x+5} & x 
eq 1 \ \ -\frac{1}{3} & x = 1 \end{cases}$$

Our mission is to find the derivative $f'(1)$.

### The Trap of the Piecewise Definition

Many students, upon seeing this, immediately reach for the quotient rule. They want to differentiate the expression and plug in $x=1$.

But stop! If you try to evaluate that expression at $x=1$, you get the indeterminate form $\frac{0}{0}$.

The function is defined differently at $x=1$ for a reason. This is a classic case where the formula for $x 
eq 1$ is just a mask for a simpler function, and the value at $x=1$ is the true anchor.

We cannot use the standard differentiation rules on the expression itself because the expression is undefined at the very point we are interested in. Instead, we must return to the soul of calculus: the **First Principle of Derivatives**.

### The Algebraic Surgery

Before we dive into the limit, let's simplify our life. Look at the denominator: $2x^2 - 7x + 5$.

Splitting the middle term, we get $2x^2 - 2x - 5x + 5$, which factors beautifully into $(x-1)(2x-5)$.

Now, look at our function for $x 
eq 1$:

$$f(x) = \frac{x-1}{(x-1)(2x-5)}$$

The $(x-1)$ terms cancel out! For all points except $x=1$, our function is simply $f(x) = \frac{1}{2x-5}$. This is the same curve, just without the hole at $x=1$.

### The First Principle

Now, we use the definition:

$$f'(1) = \lim_{h 	o 0} \frac{f(1+h) - f(1)}{h}$$

We know $f(1) = -\frac{1}{3}$. To find $f(1+h)$, we use our simplified expression:

$$f(1+h) = \frac{1}{2(1+h)-5} = \frac{1}{2h-3}$$

Now, let's assemble the limit:

$$f'(1) = \lim_{h 	o 0} \frac{\frac{1}{2h-3} - (-\frac{1}{3})}{h} = \lim_{h 	o 0} \frac{\frac{1}{2h-3} + \frac{1}{3}}{h}$$

### The Moment of Truth

To solve this, we need a common denominator for the numerator, which is $3(2h-3)$. The numerator becomes:

$$\frac{3 + (2h-3)}{3(2h-3)} = \frac{2h}{3(2h-3)}$$

Now, divide by $h$:

$$f'(1) = \lim_{h 	o 0} \frac{2h}{3h(2h-3)}$$

See the magic? The $h$ in the numerator and the $h$ in the denominator cancel out! We are left with:

$$f'(1) = \lim_{h 	o 0} \frac{2}{3(2h-3)}$$

Now, we can safely substitute $h=0$. The result is:

$$\frac{2}{3(0-3)} = \frac{2}{-9} = -\frac{2}{9}$$

### Conclusion

We have arrived at our destination: the slope of the tangent at $x=1$ is **$-\frac{2}{9}$**.

This problem teaches us that in calculus, as in life, sometimes you have to look past the surface complexity to find the simple, elegant truth underneath. Keep practicing, keep questioning, and never lose your wonder for the math!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Find the derivative of $f (x) = {\frac{x - 1}{2 x ^{2} - 7 x + 5} - \frac{1}{3} when x \neq = 1 when x = 1$ at $x = 1$ .

Visualized Solution

Visualizing the Piecewise Function

Factorizing the Denominator

Simplifying $f (x)$ for $x \neq = 1$

The Derivative Definition

Calculating $f (1 + h)$

Setting up the Limit

Simplifying the Numerator

Cancelling the Indeterminate Form

Final Evaluation

The Sigma Insight: Differentiability of a Function

The Beauty of the Piecewise Puzzle

The Trap of the Piecewise Definition

The Algebraic Surgery

The First Principle

The Moment of Truth

Conclusion

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the derivative of f(x)={2x2−7x+5x−1​−31​​when x=1when x=1​ at x=1.