How do you differentiate #g(x) = sqrt(1-x)cosx# using the product rule?

1 Answer
Jan 16, 2016

Answer:

#g'(x)=((2x-2)sinx-cosx)/(2sqrt(1-x))#

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)#, chain rule will have to be used. Treat the square root as a #1"/"2# power.

#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)#, you have to know the derivatives of simple trigonometric functions.

#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))#