# DIFFERENTIABILITY EXAMPLES

Example 1 : If f(x + y) = f(x) + f(y) - 2xy - 1 for all x and y. If f'(0) exists and f'(0) = -sin$$\alpha$$, then find f{f'(0)}.

Solution : f'(x) = $$\displaystyle{\lim_{h \to 0}}$$ $$f(x+h) - f(x)\over h$$

= $$\displaystyle{\lim_{h \to 0}}$$ $${f(x)+f(h)-2xh-1} - f(x)\over h$$

= $$\displaystyle{\lim_{h \to 0}}$$ -2x + $$\displaystyle{\lim_{h \to 0}}$$ $$f(h)-1\over h$$ = -2x + $$\displaystyle{\lim_{h \to 0}}$$ $$f(h) - f(0)\over h$$

[Putting x = 0 = y in the given relation we find f(0) = f(0) + f(0) + 0 - 1 $$\implies$$ f(0) = 1]

$$\therefore$$   f'(x) = -2x + f'(0) = -2x - sin$$\alpha$$

$$\implies$$   f(x) = -$$x^2$$ - (sin$$\alpha$$).x + c

f(0) = - 0 - 0 + c $$\implies$$ c = 1

$$\therefore$$   f(x) = -$$x^2$$ - (sin$$\alpha$$).x + 1

so, f{f'(0)} = f(-sin$$\alpha$$) = -$$sin^2\alpha$$ + $$sin^2\alpha$$ + 1 = 1

Example 2 : Discuss the continuity and differentiability of the function y = f(x) defined parametrically; x = 2t - |t-1| and y = 2$$t^2$$ + t|t|.

Solution : Here x = 2t - |t-1| and y = 2$$t^2$$ + t|t|.

Now when t < 0;

x = 2t - {-(t-1)} = 3t - 1 and y = 2$$t^2$$ - $$t^2$$ = $$t^2$$ $$\implies$$ = y = $${1\over 9}{(x+1)}^2$$

= when 0 $$\le$$ t < 1

x = 2t - {-(t-1)} = 3t - 1 and y = 2$$t^2$$ - $$t^2$$ = 3$$t^2$$ $$\implies$$ = y = $${1\over 3}{(x+1)}^2$$

when t $$\ge$$ 1;

x = 2t - (t-1) = t + 1 and y = 2$$t^2$$ + $$t^2$$ = 3$$t^2$$ $$\implies$$ = y = 3$${(x+1)}^2$$

Thus, y = f(x) = $${1\over 9}{(x+1)}^2$$, x < -1

y = f(x) = $${1\over 3}{(x+1)}^2$$, -1$$\le$$x < 2

y = f(x) = 3$${(x+1)}^2$$, x$$\ge$$ 2

We have to check differentiability at x = -1 and 2.

Differentiabilty at x = -1; LHD = f'($$-1)^-$$ = $$\displaystyle{\lim_{h \to 0}}$$$$f(-1-h) - f(-1)\over -h$$ = $$\displaystyle{\lim_{h \to 0}}$$ $${1\over 9}(-1-h+1)^2 - 0\over -h$$ = 0

RHD = f'($$-1)^+$$ = $$\displaystyle{\lim_{h \to 0}}$$$$f(-1+h) - f(-1)\over h$$ = $$\displaystyle{\lim_{h \to 0}}$$ $${1\over 3}(-1+h+1)^2 - 0\over h$$ = 0

Hence f(x) is differentiable at x = -1

$$\implies$$     continuous at x = -1.

To check differentiability at x = 2;

LHD = f'($$2)^-$$ = $$\displaystyle{\lim_{h \to 0}}$$$${1\over 3}(2-h+1)^2 - 3\over -h$$ = 2

RHD = f'($$2)^+$$ = $$\displaystyle{\lim_{h \to 0}}$$$$3(2+h-1)^2 - 3\over h$$ = 6

Hence f(x) is not differentiable at x = 2.

But continuous at x = 2, because LHD and RHD both are finite.

f(x) is continuous for all x and differentiable for all x, except x = 2.