Question

Question:
Using the first principle,find the derivative of Sin(3x²-5)?

Answer 1

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$\tt \:  ⟼ \: Let \: f(x) = sin( {3x}^{2} - 5)$

$\tt \:  ⟼ \: Change \: x \tt \:  ⟼ \: x + h$

$\tt \:  ⟼f(x+ h) = \: sin \bigg(3 {(x + h)}^{2} - 5 \bigg)$

$\tt \:  ⟼By \: definition \: of \: first \: principle$

$\tt \:  ⟼ \: f'(x) \: = \: \lim_{ h\to0}\dfrac{f(x+ h) \: - \: f(x)}{h}$

$\tt \:  \: f'(x) \: = \: \lim_{ h\to0}\dfrac{\: sin \bigg(3 {(x + h)}^{2} - 5 \bigg) - sin( {3x}^{2} - 5)}{h}$

$\tt \:  using \: sinx - siny \: = 2 \: cos(\dfrac{x + y}{2} )sin(\dfrac{x - y}{2})$

$\tt \:  = \: \lim_{ h\to0}\dfrac{2 \times cos \bigg(\dfrac{3 {(x + h)}^{2} - 5 + {3x}^{2} - 5 }{2} \bigg) \times sin \bigg(\dfrac{3 {(x + h)}^{2} - 5 - {3x}^{2} + 5 }{2} \bigg) }{h}$

$\tt \:  = \: 2 \: \lim_{ h\to0}cos \bigg(\dfrac{3 {(x + h)}^{2} - 5 + {3x}^{2} - 5 }{2} \bigg) \times \tt \:  \lim_{ h\to0}\dfrac{sin \bigg(\dfrac{3 {(x + h)}^{2} - {3x}^{2} }{2} \bigg)}{h}$

$\tt \:  = 2 \: cos( {3x}^{2} - 5) \: \lim_{ h\to0}\dfrac{sin \bigg(\dfrac{3( \cancel{{x}^{2}} + {h}^{2} + 2hx - \cancel{{x}^{2}} ) }{2} \bigg)}{h}$

$\tt \:  = 2 cos({3x}^{2} - 5) \: \lim_{ h\to0}\dfrac{sin \bigg(\dfrac{3 {h}^{2} + 6xh}{2} \bigg) }{h}$

$\tt \:  = 2 cos({3x}^{2} - 5) \: \lim_{ h\to0}\dfrac{sin \bigg(\dfrac{3 {h}^{2} + 6xh}{2} \bigg) }{\bigg(\dfrac{3 {h}^{2} + 6xh}{2} \bigg)} \times \dfrac{\bigg(\dfrac{3 {h}^{2} + 6xh}{2} \bigg)}{h}$

$\bf \:  \bigg(\because \: \tt \:  \: \lim_{ x\to0}\dfrac{sinx}{x} = \: 1 \bigg)$

$\tt \:  = 2 cos({3x}^{2} - 5) \: \lim_{ h\to0} \: 1 \times \dfrac{\bigg(\dfrac{3 {h} + 6x}{2} \bigg) \times \cancel h }{ \cancel h}$

$\tt \:  = \cancel2 cos({3x}^{2} - 5) \: \times \dfrac{6x}{ \cancel2}$

$\tt \:  = 6 \: x \: cos({3x}^{2} - 5) \:$

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