Question

15. cosec x cot x

Answer 1

Consider the function,

f( x )=cosecxcotx

According to the first principle, the derivative of the function is,

f ′ ( x )= lim h→0 f( x+h )−f( x ) h

The given function can be written as,

f( x )= f 1 ( x ) f 2 ( x )

The Leibnitz Product rule of derivative to find the derivative of the function is,

f ′ ( x )= f 1 ( x ) f ′ 2 ( x )+ f 2 ( x ) f ′ 1 ( x ) (1)

On solving the value of f 2 ′ ( x ), we get,

f ′ 2 ( x )= lim h→0 cot( x+h )−cotx h = lim h→0 1 h [ cos( x+h ) sin( x+h ) − cosx sinx ] = lim h→0 1 h [ sinxcos( x+h )−cosxsin( x+h ) sinxsin( x+h ) ]

From the trigonometric identity, we know that,

sin( A−B )=sinAcosB−cosAsinB

The derivative of the function is,

f ′ 2 ( x )= lim h→0 1 h sin( x−x−h ) sinxsin( x+h ) = lim h→0 1 h sin( −h ) sinxsin( x+h ) =− lim h→0 1 h sinh sinxsin( x+h ) =− lim h→0 sinh h lim h→0 1 sinxsin( x+h )

Apply the limits,

f ′ 2 ( x )=−( 1 ) 1 sinxsin( x+0 ) = −1 sinxsinx =−cose c 2 x

Now,

f ′ 1 ( x )= lim h→0 [ cosec( x+h )−cosecx ] h = lim h→0 1 h [ 1 sin( x+h ) − 1 sinx ] = lim h→0 1 h [ sinx−sin( x+h ) sinxsin( x+h ) ]

From the trigonometric identity, we know that,

sinC−sinD=2cos C+D 2 sin C−D 2

The derivative of the function is,

f ′ 1 ( x )= lim h→0 1 h [ 2cos( x+x+h 2 )sin( x−x−h 2 ) sinxsin( x+h ) ] = lim h→0 1 h [ 2cos( 2x+h 2 )sin( −h 2 ) sinxsin( x+h ) ] = lim h→0 −1 h 2 [ cos( 2x+h 2 ) sin( x+h )sinx sin( h 2 ) ]

Simplify the function by applying limits,

f ′ 1 ( x )= lim h→0 −cos( 2x+h 2 ) sin( x+h )sinx lim h→0 sin( h 2 ) h 2 = −cos( 2x 2 ) sinxsin( x+0 ) ( 1 ) = −cosx sinx ( cosecx ) =−cosecxcotx

Put the value of f ′ 1 ( x ) and f ′ 2 ( x ) in equation (1),

f ′ ( x )=cosecx( − cosec 2 x )+cotx( −cosecxcotx ) =− cosec 3 x−cosecx cot 2 x

Thus, the derivative of cosecxcotx is − cosec 3 x−cosecx cot 2 x.