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

Question

Emma Aldridge · Accepted Answer

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

According to the product rule,

#d/dx[f(x)h(x)]=f'(x)h(x)+h'(x)f(x)#

In this case, we have

#f(x)=sqrt(1-x)# #h(x)=cosx#

First, ascertain each of these functions' respective derivatives.

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#

Re-enter these into the expression of the product rule.

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

This is easier to understand:

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

Axel Alvarez · Answer

undefined To differentiate $ g(x) = \sqrt{1-x} \cos x $ using the product rule, you would follow these steps:

1. Identify the two functions: $ f(x) = \sqrt{1-x} $ and $ h(x) = \cos x $.
2. Apply the product rule formula: $ (f \cdot h)' = f' \cdot h + f \cdot h' $.
3. Find the derivatives of $ f(x) $ and $ h(x) $.
   - $ f'(x) = -\frac{1}{2\sqrt{1-x}} $ (using the chain rule)
   - $ h'(x) = -\sin x $ (derivative of $ \cos x $)
4. Substitute the derivatives into the product rule formula.
   - $ g'(x) = -\frac{1}{2\sqrt{1-x}} \cdot \cos x + \sqrt{1-x} \cdot (-\sin x) $
5. Simplify the expression if necessary.