Finding derivative of sin⁡2x3 using only definition of derivative

Let's use a3−b3=(a−b)(a2+ab+b2) for a=sin⁡2(x+h)3 and b=sin⁡2x3. Assume that b≠0. This implies that a2+ab+b2≠0 and we can divide:sin⁡(2x+2h)3−sin⁡2x3h=sin⁡(2x+2h)−sin⁡2xh⋅1a2+ab+b2Now using the identity sin2⁡(x+h)−sin⁡2x=2sin⁡hcos⁡(2x+h), continuity of sin and cos andlimh→0sin⁡hh=1we get that the derivative is:2cos⁡2x3sin2/3⁡2x

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iocasq4

Answered question

2022-01-25

Finding derivative of $\sqrt[3]{\sin 2 x}$ using only definition of derivative

Answer & Explanation

helsinka04

Beginner2022-01-26Added 11 answers

Let's use

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

for

a = \sqrt[3]{\sin 2 (x + h)}

and

b = \sqrt[3]{\sin 2 x}

. Assume that

b \neq 0

. This implies that

a^{2} + a b + b^{2} \neq 0

and we can divide:

\frac{\sqrt[3]{\sin (2 x + 2 h)} - \sqrt[3]{\sin 2 x}}{h} = \frac{\sin (2 x + 2 h) - \sin 2 x}{h} \cdot \frac{1}{a^{2} + a b + b^{2}}

Now using the identity

\sin^{2} (x + h) - \sin 2 x = 2 \sin h \cos (2 x + h)

, continuity of

\sin

and

\cos

and

lim_{h \to 0} \frac{\sin h}{h} = 1

we get that the derivative is:

\frac{2 \cos 2 x}{3 \sin^{2 / 3} 2 x}

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