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 Lucky f(x)=Sin(x), by first principle its f'(x)...show me how to solve such problem.

Hi Lucky,

To develop the derivative of the sine function from first principles requires that yuo show that

$\lim_{h \rightarrow 0} \frac{\sin(h)}{h} = 1.$

I am not going to prove this for you, I suggest that you look in a calculus textbook for a proof.

From first principles

$f^\prime(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}.$

For $f(x) = \sin(x)$ this becomes

$f^\prime(x) = \lim_{h \rightarrow 0} \left( \frac{\sin(x + h) - \sin(x)}{h} \right) .$

Use of the fact that $\sin(a + b) = \sin(a) \cos(b) + \cos(a) \sin(b)$ and some simplification yields

$f^\prime(x) = \lim_{h \rightarrow 0} \left( \sin(x) \frac{\cos(h) - 1}{h} + \cos(x) \frac{\sin(h)}{h} \right).$

Since we know that

$\lim_{h \rightarrow 0} \frac{\sin(h)}{h} = 1.$

it remains to evlauate

$\lim_{h \rightarrow 0} \frac{\cos(h) - 1}{h}.$

Multiplying the numerator and denominator by $\cos(h) + 1$ and using the fact that $\sin^2(h) = 1 - \cos^2(h)$ this becomes

$\lim_{h \rightarrow 0} \frac{\cos(h) - 1}{h} = \sin(x) \; \lim_{h \rightarrow 0} \frac{- \sin^2(h)}{h \cos(h) + 1}$

which simplifies to

$\lim_{h \rightarrow 0} \frac{\cos(h) - 1}{h} = -\lim_{h \rightarrow 0} \frac{\sin(h)}{h} ) \times \lim_{h \rightarrow 0}\frac{\sin(h)}{\cos(h) + 1} = -1 \times \left( \frac{0}{1 + 1} \right) = 0.$

Thus

$f^\prime(x) = \lim_{h \rightarrow 0} \sin(x) \frac{\cos(h) - 1}{h} + \cos(x) \frac{\sin(h)}{h} = 0 + \cos(x) \times 1 = \cos(x).$

Harley

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.