Find the derivative : x2·cosπ4sinx
Solution:
The quotient rule of differentiation is,
uv'=u'v-vu'v2
In the given function, u=x2,v=sinx
Lety=cosπ4x2sinx∴dydx=cosπ4sinx·ddxx2-x2·ddxsinxsin2x⇒dydx=cosπ4sinx·2x-x2·cosxsin2x⇒dydx=122x·sinx-x2·cosxsin2x
Hence, the required expression is dydx=122x·sinx-x2·cosxsin2x.