If the tangent to the curve, y = f(x) = x log_e x, (x 0) at a point (c, f(c)) is parallel to the l

Q: Wait, why do we equate the derivative to the slope of the secant line?

Great question! The problem asks for a tangent that is parallel to the secant line. In coordinate geometry, two lines are parallel if and only if their slopes are equal. The derivative $f'(c)$ gives us the slope of the tangent at any point $c$, and the slope of the secant line is constant. By equating them, we are finding the exact point where the curve's steepness matches the slope of the secant.

Q: I got stuck on the derivative of $x \log_e x$. Is it just $\log_e x$?

Not quite! Remember that $x \log_e x$ is a product of two functions: $u(x) = x$ and $v(x) = \log_e x$. You must use the Product Rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$. When you apply this, you get $x \cdot (\frac{1}{x}) + (\log_e x) \cdot 1$, which simplifies to $1 + \log_e x$. Always be careful with products!

Q: How do I simplify $\frac{e}{e-1} - 1$?

Think of $1$ as $\frac{e-1}{e-1}$. So, the expression becomes $\frac{e}{e-1} - \frac{e-1}{e-1}$. When you subtract the numerators, you get $e - (e - 1)$, which is $e - e + 1 = 1$. Thus, the whole expression simplifies beautifully to $\frac{1}{e-1}$.

Question

If the tangent to the curve, $y = f(x) = x \log_e x$, $(x > 0)$ at a point $(c, f(c))$ is parallel to the line-segment joining the points $(1,0)$ and $(e, e)$, then $c$ is equal to :

Options:

(A) $\frac{1}{e-1}$

(B) $e^{\frac{e-1}{e}}$

(C) $e^{\frac{1}{e}}$

(D) $e^{\frac{1}{e-1}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a path defined by the function $y = x \log_e x$. This curve represents a dynamic relationship between $x$ and its logarithm.

We are given two anchors, point $A(1,0)$ and point $B(e,e)$. These points define a secant line, a straight path cutting through the curve.

Our mission is to find a point $c$ on this curve where the tangent line is perfectly parallel to this secant. This is the essence of calculus: finding the moment where the instantaneous rate of change matches the average rate of change.

### The Slope of the Shortcut

First, let us calculate the slope of our secant line. The slope $m$ of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

Substituting our points $A(1,0)$ and $B(e,e)$, we get:

$$m = \frac{e - 0}{e - 1} = \frac{e}{e - 1}$$

This value represents the constant steepness of our secant line.

### The Derivative as a Slope Machine

Now, we turn to the curve itself. To find the slope of the tangent at any point $c$, we need the derivative $f'(x)$.

Our function is $f(x) = x \log_e x$. Since this is a product of two functions, we must employ the Product Rule:

$$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$$

Applying this, we get:

$$f'(x) = x \cdot \frac{d}{dx}(\log_e x) + \log_e x \cdot \frac{d}{dx}(x)$$

Since the derivative of $\log_e x$ is $\frac{1}{x}$ and the derivative of $x$ is $1$, this simplifies elegantly to:

$$f'(x) = x \cdot \frac{1}{x} + \log_e x \cdot 1 = 1 + \log_e x$$

This is our slope machine.

### The Bridge Between Geometry and Calculus

We are looking for a point $c$ where the tangent is parallel to the secant. This means the slope of the tangent at $c$, which is $f'(c) = 1 + \log_e c$, must equal the slope of the secant, $\frac{e}{e - 1}$.

So, we set up the equation:

$$1 + \log_e c = \frac{e}{e - 1}$$

Now, we solve for $c$. Subtracting $1$ from both sides, we get:

$$\log_e c = \frac{e}{e - 1} - 1$$

To simplify the right side, we find a common denominator:

$$\log_e c = \frac{e - (e - 1)}{e - 1} = \frac{e - e + 1}{e - 1} = \frac{1}{e - 1}$$

### The Final Revelation

We have arrived at $\log_e c = \frac{1}{e - 1}$. To isolate $c$, we convert this logarithmic form into its exponential counterpart.

By the definition of logarithms, if $\log_e c = y$, then $c = e^y$. Therefore, the final coordinate is:

**$$c = e^{\frac{1}{e - 1}}$$**

This is the point on the curve where the tangent is perfectly parallel to the secant line. It is a beautiful result, born from the marriage of algebraic manipulation and the fundamental principles of calculus.

Animated Solution for Mathematics - Differentiation: If the tangent to the curve, $y = f (x) = x lo g_{e} x$ , $(x > 0)$ at a point $(c, f (c))$ is parallel to the line-segment joining the points $(1, 0)$ and $(e, e)$ , then $c$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Curve and Points

The Secant Line

Slope of the Secant Line

The Tangent Line Condition

Differentiating the Function

Applying the Product Rule

Slope of Tangent at $x = c$

Equating the Slopes

Isolating $lo g_{e} c$

Simplifying the Expression

Finding the Final Value of $c$

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Slope of the Shortcut

The Derivative as a Slope Machine

The Bridge Between Geometry and Calculus

The Final Revelation

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