If f(x) = sin(u), u = v - (1/v^2), v = ln(x), then what is f’(e)?

To find \( f'(e) \), we need to use the chain rule to differentiate the given function step by step.

Given:
\[
f(x) = \sin(u)
\]
\[
u = v - \frac{1}{v^2}
\]
\[
v = \ln(x)
\]

### Step 1: Differentiate \( f(x) = \sin(u) \) with respect to \( u \)
\[
\frac{df}{du} = \cos(u)
\]

### Step 2: Differentiate \( u = v - \frac{1}{v^2} \) with respect to \( v \)
\[
\frac{du}{dv} = 1 + \frac{2}{v^3}
\]

### Step 3: Differentiate \( v = \ln(x) \) with respect to \( x \)
\[
\frac{dv}{dx} = \frac{1}{x}
\]

### Step 4: Apply the chain rule to find \( \frac{df}{dx} \)
\[
\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}
\]
Substituting the derivatives calculated:
\[
\frac{df}{dx} = \cos(u) \cdot \left(1 + \frac{2}{v^3}\right) \cdot \frac{1}{x}
\]

### Step 5: Evaluate \( f'(e) \)
First, we need to find \( v \) at \( x = e \):
\[
v = \ln(e) = 1
\]
Next, find \( u \) at \( v = 1 \):
\[
u = 1 - \frac{1}{1^2} = 1 - 1 = 0
\]
Finally, substitute \( u = 0 \), \( v = 1 \), and \( x = e \) into the derivative:
\[
f'(e) = \cos(0) \cdot \left(1 + \frac{2}{1^3}\right) \cdot \frac{1}{e}
\]
\[
f'(e) = 1 \cdot (1 + 2) \cdot \frac{1}{e} = 3 \cdot \frac{1}{e} = \frac{3}{e}
\]

So, \( f'(e) = \frac{3}{e} \).

Would you like further details or have any questions?

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1. What is the chain rule, and how is it applied in differentiation?
2. How does one differentiate trigonometric functions?
3. Can the chain rule be used for implicit differentiation?
4. What are common mistakes to avoid when applying the chain rule?
5. How does natural logarithm differentiation work?

**Tip:** Always check intermediate values, such as \( v \) and \( u \) here, to avoid errors when evaluating the derivative at specific points.

Learn how to differentiate f(x) = sin(ln(x) - 1/x^2) using the chain rule step-by-step. Calculate f'(e) and understand the application of trigonometric functions and natural logarithms in calculus. Suitable for advanced high school students and anyone interested in calculus and mathematical analysis.

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