Question

please differentiate with proper methods in notebook. I've the answer i want to know the method.​

Answer 1

Answer:

• Outside inside rule of differentiation:

• dy/dx=(differentiation of outer function keep inside as it is) × (differentiation of inner function)

Chain Rule:

If y = f(U), then dy/dx is :

$\footnotesize\implies \dfrac{dy}{dx} = \dfrac{dy}{dU} . \dfrac{dU}{dx}$

By applying Chain rule :

$\footnotesize\implies y = B \cos (Kx)$

$\footnotesize\implies \dfrac{dy}{dx} = - B \sin(Kx) \dfrac{d}{dx} (Kx)$

$\footnotesize\implies \dfrac{dy}{dx} = - B \sin(Kx) K$

$\footnotesize\implies \underline{ \boxed{ \bf \red{ \dfrac{dy}{dx} = - B K \sin(Kx)}}}$

Answer 2

Appropriate Question :-

If given that " B " is a constant then differentiate the following ;

• y = B Cos ( kx )

Understanding How given is wrong :-

Consider y = f ( x ) , then as it is not provided that ' B ' is a constant , so the given function becomes

• y = f ( x , B )

Which we can't differentiate simply , then we have to differentiate it partially because presence of more than 1 variables as follows ;

${ \quad \leadsto { \quad \bf \dfrac{\partial y}{\partial x} = \dfrac{\partial}{\partial x} \bigg[ B Cos ( kx ) \bigg] }}$

So on again Differentiating partially both sides w.r.t.B if the RHS = t ;

${ \quad \leadsto { \quad \bf \dfrac{\partial² y}{\partial x \partial B} = \dfrac{\partial}{\partial B} ( t ) }}$

Now coming back to your question ;

To Find :-

The Derivative of the given function

Solution :-

Consider , given ;

$\quad \leadsto \sf y = B Cos ( k x )$

Put ;

• kx = u

then we can write y as ;

• y = B Cos ( u )

Now consider this ;

$\quad \leadsto \sf y = B Cos ( u )$

By chain rule of Differentiation we knows that ;

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red { \dfrac{dy}{dx} = \dfrac{dy}{du} × \dfrac{du}{dx} }}}}}}{\bigstar}$

Using this we have ;

On Differentiating both sides w.r.t.x we have ;

${ : \implies \quad { \sf \dfrac{d}{dx} ( y ) = \dfrac{d}{du} \bigg[ B Cos ( kx ) \bigg] × \dfrac{d}{dx} ( u ) }}$

As " B " is a constant . So ;

${ : \implies \quad { \sf \dfrac{dy}{dx} = B . \dfrac{d}{du} ( Cos \: u ) × \dfrac{d}{dx} ( kx ) }}$

We knows that ;

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red { \dfrac{d}{dx} ( Cos \: x ) = - Sin \: x }}}}}}{\bigstar}$

Using this we have ;

Also " k " is a constant , so we can take it outside the differential operator ;

${ : \implies \quad { \sf \dfrac{dy}{dx} = B × - Sin (u) × k × \dfrac{d}{dx} ( x ) }}$

${ : \implies \quad { \sf \dfrac{dy}{dx} = B × - Sin (u) × k × 1 }}$

Put value of " u " ;

${ : \implies \quad { \sf \dfrac{dy}{dx} = k × B × - Sin (kx) }}$

${ : \implies \quad { \bf \dfrac{dy}{dx} = - B k Sin ( kx ) }}$

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red { \therefore \dfrac{dy}{dx} = - B k \: Sin ( kx ) }}}}}}{\bigstar}$