How do you differentiate #f(x)=x^3sqrt(x-2)sinx# using the product rule?

1 Answer
Mar 8, 2018

Answer:

#f'(x)=3x^2sqrt(x-2)sinx+(x^3sinx)/(2sqrt(x-2))+x^3sqrt(x-2)cosx#

Explanation:

If #f(x)=g(x)h(x)j(x)#, then #f'(x)=g'(x)h(x)j(x)+g(x)h'(x)j(x)+g(x)h(x)j'(x)#

#g(x)=x^3#
#g'(x)=3x^2#

#h(x)=sqrt(x-2)=(x-2)^(1/2)#
#h'(x)=1/2*(x-2)^(-1/2)*d/dx[x-2]#
#color(white)(h'(x))=(x-2)^(-1/2)/2*1#
#color(white)(h'(x))=(x-2)^(-1/2)/2#
#color(white)(h'(x))=1/(2sqrt(x-2))#

#j(x)=sinx#
#j'(x)=cosx#

#f'(x)=3x^2sqrt(x-2)sinx+x^3 1/(2sqrt(x-2))sinx+x^3sqrt(x-2)cosx#

#f'(x)=3x^2sqrt(x-2)sinx+(x^3sinx)/(2sqrt(x-2))+x^3sqrt(x-2)cosx#