How do you differentiate #f(x)=xsqrt(abs(x-1)# using the product rule?

1 Answer
Sep 21, 2016

Answer:

Please see the explanation section below.

Explanation:

Because it involves the absolute value function, we need to first analyze the function.

Note that #abs(x-1) = {(x-1," if ",x-1 >= 0),(1-x," if ",x-1<0):}#.

So we have

#f(x) = {(xsqrt(x-1)," if ",x > 1),(xsqrt(1-x)," if ",x < 1):}#.

Now we can differentiate using #d/dx (sqrtu) = 1/(2sqrtu) (du)/dx#.

#d/dx(xsqrt(x-1)) = (1)sqrt(x-1)+x 1/(2sqrt(x-1))*1#

# = (2x-2+x)/(2sqrt(x-1)) = (3x-2)/(2sqrt(x-1))#

and

#d/dx(xsqrt(1-x)) = (1)sqrt(1-x)+x 1/(2sqrt(1-x))*(-1)#

# = (2-2x-x)/(2sqrt(1-x)) = (2-3x)/(2sqrt(1-x))#.

Putting these together, we have

#f'(x) = {((3x-2)/(2sqrt(x-1))," if ",x > 1),((2-3x)/(2sqrt(1-x))," if ",x < 1) :}#.

Additional comments

We can write the derivative of #absx# as #{(1," if " x > 0),(-1," if "x < 0):}#

Or, we can use #absx/x# (or its reciprocal).

With this notation, we get

#d/dx (sqrtabs(x-1)) = 1/(2sqrt abs(x-1))* d/dx(abs(x-1))#

# = 1/(2sqrt(abs(x-1))) abs(x-1)/(x-1)#.

We can write #f'(x)# as

#f'(x) = (3x-2)/(2sqrt abs(x-1)) abs(x-1)/(x-1)#.