Question

19. sin"x

Answer 1

Let the given function be,

f( x )= sin n x

When n=1, the function becomes,

f( x )=sinx

Differentiate the above function with respect to x,

f ′ ( x )=cosx

When n=2, the function becomes,

f( x )= sin 2 x

Differentiate the function with respect to x by Leibnitz Product rule,

f ′ ( x )=sinx d dx ( sinx )+sinx d dx ( sinx ) =sinxcosx+sinxcosx =2sinxcosx

When n=3, the function becomes,

f( x )= sin 3 x

Differentiate the function with respect to x by Leibnitz Product rule,

f ′ ( x )= sin 2 x d dx sinx+sinx d dx ( sin 2 x ) = sin 2 xcosx+sinx( 2sinxcosx ) = sin 2 xcosx+2 sin 2 xcosx =3 sin 2 xcosx

It is assumed that d dx ( sin n x )=n sin n−1 xcosx.

Let the assumption is true for n=k, so,

d dx ( sin k x )=k sin k−1 xcosx

Now, consider that the assumption is true for n=k+1, so,

d dx ( sin k+1 x )= d dx ( sinx sin k x )

Differentiate the function with respect to x by Leibnitz Product rule,

d dx ( sinx sin k x )= sin k x d dx ( sinx )+sinx d dx ( sin k x ) =cosx sin k x+sinx( k sin k−1 xcosx ) =cosx sin k x+k sin k xcosx =( k+1 ) sin k xcosx

This shows that our assumption is true for n=k+1.

Therefore, from the mathematical induction, d dx ( sin n x )=n sin n−1 xcosx.