Q.167. Limit and Differentiation.

Question 167.
Find f '(x) given that
(1) f (x + y) = f (x) + f (y) + x^2y + xy^2 and
(2) lim (x → 0) f (x) / x = 1.

Answer 167.
f '(x)
= lim (h → 0) [f (x + h) - f (x)] / h
= lim (h → 0) [f (x) + f (h) + hx^2 + xh^2 - f (x)] / h
= lim (h → 0) [f (h) + hx^2 + xh^2] / h
= lim (h → 0) f (h) / h + lim (h → 0) (x^2 + xh)
= 1 + x^2.
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Verification:
Integrating,
f (x) = x + (1/3)x^3 + c
lim (x → 0) f (x) / x
= lim (x → 0) (1/x) (x + x^3/3)
= lim (x → 0) 1 + x^2/3
= 1 (as given)

f (x + y)
= x + y + (1/3) (x + y)^3 + c
= x + (1/3)x^3 + y + (1/3)y^3 + yx^2 + xy^2 + c
= f (x) + f(y) + yx^2 + xy^2 ... [c can be given any value]

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