Question

Differentiate y=Asin (wt-kx)

Answer 1

It is given that :- y = Asin (wt-kx)

We have to differentiate the given function with respect to x.

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d \bigg(A \: sin (wt-kx) \bigg)}{dx}$

$\rm \longrightarrow \dfrac{dy}{dx} = A \times \dfrac{d \bigg( \: sin (wt-kx) \bigg)}{dx}$

$\rm \longrightarrow \dfrac{dy}{dx} = A \times \: cos(wt-kx) \times \dfrac{d \bigg(wt-kx \bigg)}{dx}$

$\rm \longrightarrow \dfrac{dy}{dx} = A \times \: cos(wt-kx) \times (0-k)$

$\rm \longrightarrow \dfrac{dy}{dx} = - kA \: cos(wt-kx)$

Answer :- -kA cos(wt-kx)

Additional Information :-

d(e^x)/dx = e^x

d(x^n)/dx = n x^(n-1)

d(ln x)/dx = 1/x

d(sin x)/dx = cos x

d(cos x)/dx = - sin x

d(tan x)/dx = sec² x

d(sec x)/dx = tan x * sec x

d(cot x)/dx = - cosec²x

d(cosec x)/dx = - cosec x * cot x

Answer 2

Answer:

$\green{ \sf \: \frac{dy}{dx} = - Ak \: cos(wt - kx)}$

Explanation:

Given,

$\sf \: y = Asin(wt - kx)$

we have to differenciate it.

But, at first we have to know some formula.

$\pink{ \odot \: \sf \frac{d \: sin \: x}{dx} = cos \: x } \\ \\ \purple{ \odot \: \sf \: \frac{d \: (constant)}{dx} = 0} \\ \\ \red{ \odot \: \sf \: \frac{d \: {x}^{n} }{dx} = n {x}^{n - 1} }$

Solution :

$\sf \: y = Asin(wt - kx) \\ \therefore \sf \: \frac{dy}{dx} = \frac{d \: \{ Asin(wt - kx)\}}{dx} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \sf = Acos(wt - kx) \times \frac{d \: (wt - kx)}{dx} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \sf = Acos( wt- kx) \times {\huge{ (}} \frac{d \: wt}{dx} - \frac{d \: kx}{dx} { \huge {)}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \sf = Acos(wt - kx) \times (0 - k) \\ \\ \sf \therefore \: \frac{dy}{dx} \sf \: = - Ak \: cos(wt - kx)$