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

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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.

Practice these given differentiability examples to test your knowledge on concepts of differentiability.