Question

Find d/dx when the (sin omega t) is given

Answer 1

Answer:

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

CALCULATION:

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

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

dx

d

(y) .

y is differentiated with respect to x as follows-

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

dx

d

(y)=

dx

d

[Asin(ωt−kx)]

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

dx

d

sin(ωt−kx)

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

Putting chain rule on the above expression, we get

\frac{d}{dx}(y)

dx

d

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

d(ωt−kx)

d

sin(ωt−kx)×

dx

d

(ωt−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(ωt−kx)×[

dx

d

(ωt)−k

dx

d

(x)]

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

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