Question

derivation of sin^nx sin nx​

Answer 1

ANSWER:

$n\ sin^{n-1}x(sin \ nx \ cos \ x + cos\ nx\ sin \ x)$

GIVEN:

$\sin^nx \sin nx$

TO DIFFERENTIATE:

$\sin^nx \sin nx$

FORMULAE:

$\dfrac{d}{dx} (uv )= uv' + vu'$

$\dfrac{d}{dx} x^n = nx^{n-1}$

$\dfrac{d}{dx} \sin x = \cos x$

$\dfrac{d}{dx} kx = k$

EXPLANATION:

${\sin nx \dfrac{d}{dx} \sin^nx + \sin^nx\dfrac{d}{dx} \sin nx }$

${ \sin nx \left( n \sin^{n-1} x\right) \dfrac{d}{dx} \sin x+ \sin^nx\cos nx \dfrac{d}{dx} nx}$

${ n\sin nx \: \sin^{n-1} x \cos x+n \sin^nx\cos nx }$

$n\ sin^{n-1}x(sin \ nx \ cos \ x + cos\ nx\ sin \ x)$

SOME MORE FORMULAE:

${\boxed{\begin{minipage}{5cm} \bold{ \dfrac{d}{dx} (constant) = 0}\\ \\ \bold{ \dfrac{d}{dx}cos \ x=-sin \ x }\\ \\ \bold{ \dfrac{d}{dx}tan \ x=sec^2 \ x }\\\\\bold{ \dfrac{d}{dx}sec \ x=sec\ x\ tan\ x }\\ \\ \bold{ \dfrac{d}{dx}cosec \ x= - cosec \ x\ cot \ x }\\ \\ \bold{ \dfrac{d}{dx}cot \ x=-cosec^2\ x }\end{minipage}}$