Another Derivative Question

[Jason76]

What is it's derivative?

\(\displaystyle f(x) = (x - \sqrt{x})(x + \sqrt{x})\)

At first glance, some might say the product rule. But acually this can be de-factored into \(\displaystyle x^{2} - x(\sqrt{x})\)

\(\displaystyle f(x) = (x - \sqrt{x})(x + \sqrt{x})\)

\(\displaystyle f(x) = x^{2} - x(\sqrt{x})\)

\(\displaystyle f(x) = x^{2} - x(x^{1/2})\)

\(\displaystyle f(x) = x^{2} - x^{3/2}\) - So now would you just use the power rule?

 

[Bob Brown MSEE]

What is it's derivative?

\(\displaystyle f(x) = (x - \sqrt{x})(x + \sqrt{x})\)

At first glance, some might say the product rule. But acually this can be de-factored into \(\displaystyle x^{2} - x(\sqrt{x})\)

\(\displaystyle f(x) = (x - \sqrt{x})(x + \sqrt{x})\)

\(\displaystyle f(x) = x^{2} - x(\sqrt{x})\) <=== check this

\(\displaystyle f(x) = x^{2} - x(x^{1/2})\)

\(\displaystyle f(x) = x^{2} - x^{3/2}\) - So now would you just use the power rule?

above

 

[Jason76]

above

:idea: \(\displaystyle \sqrt{x}*\sqrt{x} = x\)

\(\displaystyle f(x) = (x - \sqrt{x})(x + \sqrt{x})\)

\(\displaystyle f(x) = (x - \sqrt{x})(x + \sqrt{x})\)

\(\displaystyle f(x) = x^{2} - x\)

\(\displaystyle f'(x) = \dfrac{d}{dx}[x^{2}] - \dfrac{d}{dx}[x]\)

\(\displaystyle f'(x) = 2x - 1\) - Answer

Also is my notation using \(\displaystyle \dfrac{d}{dx}\) ok in regards to \(\displaystyle f'(x)\)?

 

[Bob Brown MSEE]

perfect -- good work
You might add x > or = 0
because f(x) is undefined for real numbers otherwise,

 

[Jason76]

perfect -- good work
You might add x > or = 0
because f(x) is undefined for real numbers otherwise,

Perhaps the online homework would want the product rule, since the restrictions cause problems.

 

[Deleted member 4993]

Perhaps the online homework would want the product rule, since the restrictions cause problems.

No matter which rule you use, the constraint x ≥ 0 will remain (because √x is not in the real range for x < 0)