Question

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


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Answer 1

Answer:

Leibntiz theorem applied when two function in Product form.

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

• According to the question –

\implies f(x) = \sin(2x) \\

• We know that –

sin(2x) = 2 \sin(x) \cos(x) \\

• So –

\\ \bf \implies f(x) = 2 \sin(x) \cos(x) \\

Answer 2

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) –

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

• According to the question –

$\\ \bf \implies f(x) = \sin(2x)$

• We know that –

$\\ \bf \implies \sin(2x) = 2 \sin(x) \cos(x)$

• So –

$\\ \bf \implies f(x) = 2 \sin(x) \cos(x)$

• Let's find first derivative –

$\\ \bf \implies \dfrac{d \{f(x) \}}{dx} = 2 \dfrac{d\{ \sin(x). \cos(x) \}}{dx}$

$\\ \bf \implies \dfrac{d \{f(x) \}}{dx} = 2 \left \{^{1}c_{0} \dfrac{d \sin(x) }{dx}( \cos x) + \:^{1}c_{1}( \sin x) \dfrac{d \cos(x) }{dx} \right \}$

$\\ \bf \implies f'(x)= 2 \left \{ \dfrac{d \sin(x) }{dx}( \cos x) + \:( \sin x) \dfrac{d \cos(x) }{dx} \right \}$

$\\ \bf \implies f'(x)= 2 \left \{ \cos(x). \cos (x) + \sin(x) \{ - \sin(x) \} \right\}$

$\\ \bf \implies f'(x)= 2 \left \{ \cos^{2} (x) - \sin^{2} (x)\right\}$

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