How do you differentiate #g(x) = sqrt(1-x)cosx# using the product rule?
1 Answer
Jan 16, 2016
Explanation:
The product rule states that
#d/dx[f(x)h(x)]=f'(x)h(x)+h'(x)f(x)#
In this scenario, we have
#f(x)=sqrt(1-x)#
#h(x)=cosx#
First, find the derivative of each of these individual functions.
To find
#f'(x)=d/dx[(1-x)^(1/2)]=1/2(1-x)^(-1/2) * d/dx[1-x]#
#=1/(2sqrt(1-x)) * -1=-1/(2sqrt(1-x))#
To find
#h'(x)=d/dx[cos(x)]=-sinx#
Plug these back into the product rule expression.
#g'(x)=-cosx/(2sqrt(1-x))-sinxsqrt(1-x)#
This can be simplified:
#g'(x)=((2x-2)sinx-cosx)/(2sqrt(1-x))#