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

Question

Aurora Alvarado · Accepted Answer

Say #y=g(x)#

This means that:

#y=sqrt(e^x-x)*cosx#

#y=u*v#

Whereby both u and v are functions of x .

If this is the case, then:

#(dy)/(dx)=u*(dv)/(dx)+v*(du)/(dx)=g'(x)#

Now... Using implicit differentiation...

#u=sqrt(e^x-x)#

#u^2=e^x-x#

#2u*(du)/(dx)=e^x-1#

#(du)/(dx)=(e^x-1)/(2u)#

#(du)/(dx)=(e^x-1)/(2sqrt(e^x-x))#

Using normal differentiation...

#v=cosx#

#(dv)/(dx)=-sinx#

This means that:

#(dy)/(dx)=-sinxsqrt(e^x-x)+cosx*(e^x-1)/(2sqrt(e^x-x))#

Therefore:

#g'(x)=-sinxsqrt(e^x-x)+cosx*(e^x-1)/(2sqrt(e^x-x))#

Adam Adkins · Answer

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

1. Identify the two functions being multiplied: $ f(x) = \sqrt{e^x - x} $ and $ h(x) = \cos(x) $.
2. Apply the product rule formula: $ g'(x) = f'(x)h(x) + f(x)h'(x) $.
3. Differentiate each function separately:
   - For $ f(x) = \sqrt{e^x - x} $, use the chain rule to find $ f'(x) $.
   - For $ h(x) = \cos(x) $, differentiate it directly to find $ h'(x) $.
4. Plug the derivatives into the product rule formula.
5. Simplify the expression to obtain the final result.