Question

$\huge\bold\red{➡ǫᴜᴇsᴛɪᴏɴ}$
Evaluate the derivative of f(x) = sin2x using Leibnitz product rule.

✔Only mods/stars are allowed to answer​​

Answer 1

We have,

$f(x) = sin {}^{2} 2x$

On differentiating w.r.t X , we get

$f {}^{1}(x) = \frac{d}{dx} sin {}^{2} 2 x$

${f}^{1} (x)2sin \: 2x \frac{d}{dx} (sin2)$

${f}^{1} (x) = 2sin \: 2xcos2x \times (2)$

${f}^{1} (x) = 2sin \: 4x$

Hence, this is the answer.

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Answer 2

${\huge{\underline{\bf{\pink{Question}}}}}$

Evaluate the derivative of f(x) = sin2x using Leibnitz product rule.

${\huge{\underline{\bf{\blue{Answer}}}}}$

TO FIND

• Derivative of the function f(x) = sin(2x) by using Leibntiz product rule.

SOLUTION

• Leibntiz theorem applied when two function in Product form.

• If two function are 'u' & 'v' then nth Derivative of (u.v) –

$\begin{gathered} \\ \bf \implies \dfrac{d^{n}(u.v)}{dx} = \: ^{n}c_{0} \dfrac{d^{n}(u)}{dx} (v) + \: ^{n}c_{1} \dfrac{d^{n - 1}(u)}{dx} \left(\dfrac{d(v)}{dx} \right) + .. + \: ^{n}c_{n}(u) \dfrac{d^{n}(v)}{dx} \\ \end{gathered}$

• According to the question

$\begin{gathered} \\ \bf \implies f(x) = \sin(2x) \\ \end{gathered} ⟹f(x)=sin(2x)$

• We know that

$\begin{gathered} \\ \bf \implies \sin(2x) = 2 \sin(x) \cos(x) \\ \end{gathered} ⟹sin(2x)=2sin(x)cos(x)$

• So –

$\begin{gathered} \\ \bf \implies f(x) = 2 \sin(x) \cos(x) \\ \end{gathered} ⟹f(x)=2sin(x)cos(x)$

• Let's find first derivative

$\begin{gathered} \\ \bf \implies \dfrac{d \{f(x) \}}{dx} = 2 \dfrac{d\{ \sin(x). \cos(x) \}}{dx} \\ \end{gathered} ⟹ dxd{f(x)}​ =2 dxd{sin(x).cos(x)}$

$\begin{gathered} \\ \bf \implies \dfrac{d \{f(x) \}}{dx} = 2 \left \{^{1}c_{0} \dfrac{d \sin(x) }{dx}( \cos x) + \:^{1}c_{1}( \sin x) \dfrac{d \cos(x) }{dx} \right \} \\ \end{gathered} ⟹ dxd{f(x)}​ =2{ 1 c 0​ dxdsin(x)​ (cosx)+ 1 c 1​ (sinx) dxdcos(x)​ }$$\begin{gathered} \\ \bf \implies f'(x)= 2 \left \{ \dfrac{d \sin(x) }{dx}( \cos x) + \:( \sin x) \dfrac{d \cos(x) }{dx} \right \} \\ \end{gathered}$

$\begin{gathered} \\ \bf \implies f'(x)= 2 \left \{ \cos(x). \cos (x) + \sin(x) \{ - \sin(x) \} \right\} \\ \end{gathered} ⟹f ′ (x)=2{cos(x).cos(x)+sin(x){−sin(x)}}$

$\begin{gathered} \\ \bf \implies f'(x)= 2 \left \{ \cos^{2} (x) - \sin^{2} (x)\right\} \\ \end{gathered} ⟹f ′ (x)=2{cos 2 (x)−sin 2 (x)}$

$\begin{gathered} \\ \bf \implies f'(x)= 2 \left \{ \cos(2x) \right\} \: \: \: \: \: \: \: \left[\: \because \: \cos^{2} (x) - \sin^{2} (x) = \cos(2x) \right] \\ \end{gathered}$

$\begin{gathered} \\ \bf \large \implies{ \boxed{ \bf f'(x)= 2 \cos(2x)}} \\ \end{gathered} ⟹ f ′ (x)=2cos(2x)$

Thanks!