Thread: Input On A Derivative .. x(absolute value of X )

1. Excuse my ignorance .. my math is rusty very rusty Playing with the graphing calculator I put together the equation ; X( square root of X squared ). The graph looked interesting so I took it's derivative by hand and after fumbling it up a few times I worked out and justified the same derivative as the calculator ; f'(x) = 2 * absolute value of X.

The question I have is that when I change the form of the original function to ; X( absolute value of X) .. replacing the the square root of x squared with absolute value of X which should be equivalent ?? .. I can't seem to get the same derivative back again as an answer. I get instead absX + X.

Could it be simply based on the definition of square root of x ? I can only think that the derivative of the absolute value of X would be 1. Simple enough but applying the product rule to X(abs of X) leaves me with absX + X. Does the defintion of square root of X dictate that X in this case must be the abs of X and so add to equal 2(abs of X ) ?

Thanks ...   2.

3. Originally Posted by MohaveBiologist
Excuse my ignorance .. my math is rusty very rusty Playing with the graphing calculator I put together the equation ; X( square root of X squared ). The graph looked interesting so I took it's derivative by hand and after fumbling it up a few times I worked out and justified the same derivative as the calculator ; f'(x) = 2 * absolute value of X.

The question I have is that when I change the form of the original function to ; X( absolute value of X) .. replacing the the square root of x squared with absolute value of X which should be equivalent ?? .. I can't seem to get the same derivative back again as an answer. I get instead absX + X.

Could it be simply based on the definition of square root of x ? I can only think that the derivative of the absolute value of X would be 1. Simple enough but applying the product rule to X(abs of X) leaves me with absX + X. Does the defintion of square root of X dictate that X in this case must be the abs of X and so add to equal 2(abs of X ) ?
My guess is that the calculator is having trouble with taking the derivative. It is not a standard function. |x| fails to be differentiable at 0 although x|x| is differentiable everywhere. I suspect that the calculator in trying to work with |x| is attempting to use the product rule with x|x| but it does not apply at 0 where |x| fails to be differentiable.

Generally speaking, I would trust hand calculations over the results of a calculator for things like differentiation. Besides I know what the result is supposed to be when I do it by hand, and I don't know what some programmer did to handle symbolic differentiation in the calculator.  Posting Permissions
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