Question

find the derivative of √(sin 2x) using the 1st principal​

Answer 1

sin(2x)−−−−−−√sin⁡(2x)

f(x)=sin(2x)−−−−−−√f(x)=sin⁡(2x)

f(x+Δx)=sin(2x+2Δx)−−−−−−−−−−−√f(x+Δx)=sin⁡(2x+2Δx)

Now,

f′(x)=limΔx→0f(x+Δx)−f(x)Δxf′(x)=limΔx→0f(x+Δx)−f(x)Δx

=limΔx→0sin(2x+2Δx)−−−−−−−−−−−√−sin(2x)−−−−−−√

Answer 2

Step-by-step explanation:

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$find \: \: derivative \: \: of \implies \\ \\ \sqrt{ \sin(2x) }$

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

$\large \red{formula} \\ \\ \sin(2x) = 2 \sin(x) \cos(x)$

$\\ \\ \implies \frac{d}{dx}( \sqrt{ \sin(2x) } )\\ \\ \implies \frac{d}{dx} ( \sqrt{2 \sin(x) \cos(x) } ) \\ \\ \implies \frac{d}{dx} {(2 \sin(x) \cos(x) )}^{ (\frac{1}{2} )} \\ \\ \implies \frac{ \not2}{ \not2} ( \cos(x) \times ( - \sin(x) ) \\ \\ \implies - \sin(x) \cos(x)$