Question

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

Answer 1

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)`.