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Step-by-Step Solution
Step 1: Understand the Given Functional Equation
We have a differentiable function f(x) satisfying
f(x+y) = f(x) + f(y) + x\,y^2 + x^2\,y for all real x and y .
Our goal is to find f'(3) given that
\lim_{x \to 0} \frac{f(x)}{x} = 1 .
Step 2: Differentiate the Equation Partially with Respect to x
From f(x+y) = f(x) + f(y) + x\,y^2 + x^2\,y , treat y as a constant and differentiate both sides with respect to x . This yields:
f'(x+y) = f'(x) + \frac{d}{dx}[x\,y^2 + x^2\,y].
Now compute the derivative of x\,y^2 + x^2\,y with respect to x :
\frac{d}{dx}(x\,y^2) = y^2,
\frac{d}{dx}(x^2\,y) = 2\,x\,y.
So,
f'(x+y) = f'(x) + y^2 + 2\,x\,y.
Step 3: Substitute x = 0
Let x = 0 in the above equation:
f'(y) = f'(0) + y^2 + 2\cdot 0 \cdot y.
Hence,
f'(y) = f'(0) + y^2.
Thus, we get a general form for f'(x) :
f'(x) = f'(0) + x^2.
Step 4: Find f(0) from the Original Equation
Put x = 0 and y = 0 in
f(x+y) = f(x) + f(y) + x\,y^2 + x^2\,y :
f(0) = f(0) + f(0) + 0 + 0 \implies f(0) = 2f(0).
Therefore, f(0) = 0.
Step 5: Use the Limit Condition to Determine f'(0)
We are given
\lim_{x \to 0} \frac{f(x)}{x} = 1.
As x \to 0, f(x) \to 0, so the expression \frac{f(x)}{x} is of the form \frac{0}{0}. We can apply L'Hospital's Rule:
\lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{f'(x)}{1}.
Given that this limit equals 1:
\lim_{x \to 0} f'(x) = 1.
Since f'(x) = f'(0) + x^2, as x \to 0, f'(x) \to f'(0). Therefore,
f'(0) = 1.
Step 6: Substitute f'(0) = 1 into the Expression for f'(x)
We found that f'(x) = f'(0) + x^2. Substituting f'(0) = 1 gives
f'(x) = 1 + x^2.
Step 7: Find f'(3)
Finally, to get f'(3), substitute x = 3 into f'(x) = 1 + x^2:
f'(3) = 1 + 3^2 = 1 + 9 = 10.
Answer
f'(3) = 10.
