If y=f(2x+3)(3-2x) and f'(x)=sin(logx), then dydx is equal to
sin(logx)xlogx
12(3-2x)2sinlog2x+33-2x
cos(logx)xlogx
none of these
Explanation for the correct option.
Given that, y=f(2x+3)(3-2x) and f'(x)=sin(logx)
Differentiate y with respect to xusing chain rule.
dydx=f'(2x+3)(3-2x)d(2x+3)(3-2x)dx
Use quotient rule i.e; duvdx=vdudx-udvdxv2.
dydx=f'(2x+3)(3-2x)×[(3-2x)×2-(2x+3)×-2](3-2x)2=sinlog2x+33-2x×123-2x2
Hence, option B is correct.