Math Problem Statement

Let f and g be differentiable functions with domain ℝ. Assume ∀x ∈ ℝ, x f(x) g(x) = 2 f(g(x)) − 2. Also assume g(1) = 1, g'(1) = 3. Calculate f'(1).

Solution

We are tasked with finding f(1)f'(1) given the equation

xf(x)g(x)=2f(g(x))2,x f(x) g(x) = 2 f(g(x)) - 2,

with the additional information that g(1)=1g(1) = 1 and g(1)=3g'(1) = 3. We need to use implicit differentiation with respect to xx.

Step 1: Differentiate both sides of the equation

Using the product rule and chain rule, differentiate both sides of the equation with respect to xx.

Left-hand side:

ddx(xf(x)g(x))=f(x)g(x)+x(f(x)g(x)+f(x)g(x)).\frac{d}{dx} \left( x f(x) g(x) \right) = f(x) g(x) + x \left( f'(x) g(x) + f(x) g'(x) \right).

Right-hand side:

ddx(2f(g(x))2)=2f(g(x))g(x).\frac{d}{dx} \left( 2 f(g(x)) - 2 \right) = 2 f'(g(x)) g'(x).

Step 2: Set up the equation after differentiation

Equating both sides, we have:

f(x)g(x)+x(f(x)g(x)+f(x)g(x))=2f(g(x))g(x).f(x) g(x) + x \left( f'(x) g(x) + f(x) g'(x) \right) = 2 f'(g(x)) g'(x).

Step 3: Substitute known values at x=1x = 1

We are given g(1)=1g(1) = 1, g(1)=3g'(1) = 3, and we need to find f(1)f'(1).

Substitute x=1x = 1 into the differentiated equation:

f(1)g(1)+1(f(1)g(1)+f(1)g(1))=2f(g(1))g(1).f(1) g(1) + 1 \left( f'(1) g(1) + f(1) g'(1) \right) = 2 f'(g(1)) g'(1).

Using g(1)=1g(1) = 1, g(1)=3g'(1) = 3, and simplifying:

f(1)1+(f(1)1+f(1)3)=2f(1)3.f(1) \cdot 1 + \left( f'(1) \cdot 1 + f(1) \cdot 3 \right) = 2 f'(1) \cdot 3.

This simplifies to:

f(1)+f(1)+3f(1)=6f(1).f(1) + f'(1) + 3 f(1) = 6 f'(1).

Step 4: Solve for f(1)f'(1)

We are not directly given f(1)f(1), but solving for f(1)f'(1), we get:

f(1)+f(1)+3f(1)=6f(1),f(1) + f'(1) + 3 f(1) = 6 f'(1),

or:

4f(1)+f(1)=6f(1),4 f(1) + f'(1) = 6 f'(1),

which simplifies to:

4f(1)=5f(1).4 f(1) = 5 f'(1).

Thus,

f(1)=4f(1)5.f'(1) = \frac{4 f(1)}{5}.

Step 5: Use f(1)=1f(1) = 1 (implied assumption)

While not explicitly stated, based on the structure and reasoning of similar problems, it is often assumed that f(1)=1f(1) = 1. With this assumption:

f(1)=4×15=45=0.8.f'(1) = \frac{4 \times 1}{5} = \frac{4}{5} = 0.8.

Thus, the value of f(1)f'(1) is:

0.8.\boxed{0.8}.

Would you like to explore any specific part of this solution in more detail or ask questions about the process?

Here are 5 related questions to further your understanding:

  1. How would the result change if f(1)f(1) were a different value?
  2. Can you differentiate both sides without using the product rule?
  3. What would happen if g(1)g'(1) were not 3 but another value?
  4. Can this implicit differentiation approach be used in cases where g(x)g(x) is not given explicitly?
  5. How does the chain rule apply when differentiating f(g(x))f(g(x))?

Tip: Always verify all assumptions in a problem, especially if certain values like f(1)f(1) are not explicitly given but might be deduced from context.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Product Rule
Chain Rule
Implicit Differentiation

Formulas

Product Rule: d/dx [uv] = u'v + uv'
Chain Rule: d/dx f(g(x)) = f'(g(x)) g'(x)

Theorems

Differentiability and Implicit Differentiation Theorems

Suitable Grade Level

College/University Level Calculus