Question

If y= A sin(wt-kx) then the value of dy/dx is

Answer 1

may be its helpful for u

Answer 2

The correct answer to the question will be  $\ -Akcos(\omega t-kx)$

CALCULATION:

We have been given $y\ =\ Asin(\omega t-kx)$ .

We are asked to differentiate y with respect to x i.e $\frac{d}{dx}(y)$ .

y is differentiated with respect to x as follows-

                                                 $\frac{d}{dx}(y)=\ \frac{d}{dx}[Asin(\omega t-kx)]$

                                                       $=\ A\frac{d}{dx}sin(\omega t-kx)$

Here, A is taken outside as it is a constant.

Putting chain rule on the above expression, we get

                $\frac{d}{dx}(y)$ $=A\frac{d}{d(\omega t-kx)}sin(\omega t-kx)\times \frac{d}{dx}(\omega t-kx)$

Here, we are differentiating with respect to x . Hence, ωt will be considered as a constant with respect to x.

                                                      ⇒$=A\ cos(\omega t-kx)\times [\frac{d}{dx}(\omega t)-k\frac{d}{dx}(x)]$

                                                       $=A\ cos(\omega t-kx) [0-k]$

                                                       $=\ -Akcos(\omega t-kx)$  [ans]