Question

please solve with solution ​

Answer 1

Answer:

$\frac{d}{dx} ( \sin( \sqrt{2x - 3} ) = \frac{ \cos(2x - 3) }{ \sqrt{2x - 3} }$ (1)

Step-by-step explanation:

→ $\frac{d}{dx}( \sin( \sqrt{2x - 3} \:)$

$= \cos( \sqrt{2x - 3} ) \times \frac{d}{dx} ( \sqrt{2x - 3} )$ - eq.1

Here,

→ $\frac{d}{dx} ( \sqrt{2x - 3}) = \frac{1}{2 \sqrt{2x - 3} } \times \frac{d}{dx} (2x - 3)$

→ $\frac{d}{dx} (\sqrt{2x - 3} ) = \frac{1}{ \sqrt{2x - 3} }$

Now substitute this value in eq.1

So,

→ $\frac{d}{dx} ( \sin(2x - 3) ) = \frac{ \cos(2x - 3) }{ \sqrt{2x - 3} }$

Formulae used:

$1) \frac{d}{dx} ( \sin(x) ) = \cos(x)$

$2) \frac{d}{dx} ( {x}^{n} ) = n {x}^{n - 1}$