How do you differentiate #f(x)=xsinx# using the product rule?

2 Answers

hi
PRODUCT RULE SAYS
if to diffrentiate "uv"( here i will take w.r.t to x)
i.e
#d((uv))/dx#=#u*d(v)/dx#+#v*du/dx#
i.e u must know
diffrentiation of #x^n#=#nx^(n-1)#
here..,

**diffrentiation of x=(1)#x^(1-1)# =1

diffrentiation of" sinx" is "#cos x #"

NOW,
#dx.sin(x)/dx#=#(x).dsin(x)/dx#+#(sinx)(d(x)/dx)# = xcos(x)+sin(x)
=
So diffrentiation of F(X)=x sin(x)-=x cos(x)+sin(x)

Nov 12, 2017

Answer:

#f'(x)=xcosx+sinx#

Explanation:

#"given "f(x)=g(x)h(x)" then"#

#f'(x)=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"#

#g(x)=xrArrg'(x)=1#

#h(x)=sinxrArrh'(x)=cosx#

#rArrf'(x)=xcosx+sinx#