Question

Differentiate sin cube of x from first principles

Answer 1

Step-by-step explanation:

Let $y=f(x)=\sin^3(x)$

Now,

${f}^{ \prime} (x) = \lim_{h \rarr0} \frac{f(x + h) - f(x)}{h}$

$\implies {f}^{ \prime} (x) = \lim_{h \rarr0} \frac{ \sin ^{3} (x + h) - \sin^{3} (x)}{h}$

$\implies {f}^{ \prime} (x) = \lim_{h \rarr0} \frac{ \{\sin(x + h) - \sin (x) \} \{ \sin^{2} (x + h) +\sin(x + h)\sin(x) + \sin^{2} (x) \}}{h}$

$\implies {f}^{ \prime} (x) = \lim_{h \rarr0} \frac{ \{\sin(x + h) - \sin (x) \}}{h}. \lim_{h \rarr0} \{ \sin^{2} (x + h) +\sin(x + h)\sin(x) + \sin^{2} (x) \}$

$\implies {f}^{ \prime} (x) = \lim_{h \rarr0} \bigg \{\frac{ \sin(x ) \cos(h) + \cos(x) \sin(h) - \sin (x) }{h} \bigg \}. \lim_{h \rarr0} \{ \sin^{2} (x + h) +\sin(x + h)\sin(x) + \sin^{2} (x) \}$

$\implies {f}^{ \prime} (x) = \lim_{h \rarr0} \bigg \{\frac{ \sin(x ) ( \cos(h) - 1) + \cos(x) \sin(h) }{h} \bigg \}. \lim_{h \rarr0} \{ \sin^{2} (x + h) +\sin(x + h)\sin(x) + \sin^{2} (x) \}$

$\implies {f}^{ \prime} (x) = \lim_{h \rarr0} \bigg \{\frac{ \sin(x ) ( \cos(h) - 1)}{h} + \frac{\cos(x) \sin(h) }{h} \bigg \}. \{ \sin^{2} (x + 0) +\sin(x + 0)\sin(x) + \sin^{2} (x) \}$

$\implies {f}^{ \prime} (x) = \bigg \{ \lim_{h \rarr0} \frac{ \sin(x ) ( \cos(h) - 1)}{h} + \lim_{h \rarr0} \frac{\cos(x) \sin(h) }{h} \bigg \}. \{ \sin^{2} (x) +\sin(x)\sin(x) + \sin^{2} (x) \}$

$\implies {f}^{ \prime} (x) = \bigg \{ \sin(x) .\lim_{h \rarr0} \frac{ \cos(h) - 1}{h} + \cos(x). \lim_{h \rarr0} \frac{ \sin(h) }{h} \bigg \}. 3\sin^{2} (x)$

$\implies {f}^{ \prime} (x) = 3 \sin^{2} (x) . \bigg \{ \sin(x) .\lim_{h \rarr0} \frac{ \frac{d}{dh} ( \cos(h) - 1)}{ \frac{d}{dh}( h)} + \cos(x).1 \bigg \}$

$\implies {f}^{ \prime} (x) = 3 \sin^{2} (x) . \bigg \{ \sin(x) .\lim_{h \rarr0} \frac{ - \sin(h) }{1} + \cos(x) \bigg \}$

$\implies {f}^{ \prime} (x) = 3 \sin^{2} (x) . \bigg \{ - \sin(x) . \sin(0) + \cos(x) \bigg \}$

$\implies {f}^{ \prime} (x) = 3 \sin^{2} (x) . \bigg \{0 + \cos(x) \bigg \}$

$\implies {f}^{ \prime} (x) = 3 \sin^{2} (x) . \cos(x)$