Question

find the derivative of sin^n x​

Answer 1

Question :

Find the Derivatives of the $\sf\sin^{n}x$

Formula's Used :

• Genral Formula

$1)\sf\:\frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}$

$2)\sf\:\frac{d(constant)}{dx} = 0$

$3)\sf\dfrac{d(\sin\:x)}{dx}=\cos\:x$

$\sf4)\dfrac{d(\cos\:x)}{dx}=-\sin\:x$

• Chain rule

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\sf\:\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$

Solution :

We have to find the derivative of$\sf\sin^{n}x$

$\sf\:y=\sin^{n}x$

Let $\sf\sin\:x=t$,Then

$\sf\:y=(t)^n$

Now Differentiate it with respect to x, by chain rule

$\sf\implies\dfrac{dy}{dx}=\dfrac{d(t^n)}{dt}\times\dfrac{dt}{dx}$

$\sf\implies\dfrac{dy}{dx}=\dfrac{d(\sin\:x)^n}{d(\sin\:x)}\times\dfrac{d(\sin\:x)}{dx}$

$\sf\implies\dfrac{dy}{dx}=n\sin^{n-1}x\:\times\cos\:x$

It is the required solution

_______________

More Information :

• Quotient rule

Let u = f(x) and v = g(x)

Then ,

$\sf\dfrac{d}{dx}(\dfrac{u}{v})=\dfrac{v\times\frac{du}{dx}-u\times\frac{dv}{dx}}{(v)^2}$

• Product Rule

Let u = f(x) and v = g(x) , then

$\sf\dfrac{d(uv)}{dx}=u\times\dfrac{dv}{dx}+v\times\dfrac{du}{dx}$

Answer 2

Question :

Find the Derivatives of the $\sf\sin^{n}x$

Formula's Used :

• Genral Formula

$1)\sf\:\frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}$

$2)\sf\:\frac{d(constant)}{dx} = 0$

$3)\sf\dfrac{d(\sin\:x)}{dx}=\cos\:x$

$\sf4)\dfrac{d(\cos\:x)}{dx}=-\sin\:x$

• Chain rule

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\sf\:\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$

Solution :

We have to find the derivative of$\sf\sin^{n}x$

$\sf\:y=\sin^{n}x$

Let $\sf\sin\:x=t$,Then

$\sf\:y=(t)^n$

Now Differentiate it with respect to x, by chain rule

$\sf\implies\dfrac{dy}{dx}=\dfrac{d(t^n)}{dt}\times\dfrac{dt}{dx}$

$\sf\implies\dfrac{dy}{dx}=\dfrac{d(\sin\:x)^n}{d(\sin\:x)}\times\dfrac{d(\sin\:x)}{dx}$

$\sf\implies\dfrac{dy}{dx}=n\sin^{n-1}x\:\times\cos\:x$

It is the required solution

_______________

More Information :

• Quotient rule

Let u = f(x) and v = g(x)

Then ,

$\sf\dfrac{d}{dx}(\dfrac{u}{v})=\dfrac{v\times\frac{du}{dx}-u\times\frac{dv}{dx}}{(v)^2}$

• Product Rule

Let u = f(x) and v = g(x) , then

$\sf\dfrac{d(uv)}{dx}=u\times\dfrac{dv}{dx}+v\times\dfrac{du}{dx}$

$\huge\colorbox{lime}{sᴀᴋsʜɪ࿐ ❤}$