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I am asked to differentiate the following:

$$g(t) =\sqrt t (t-1)$$

The first thing I do is apply the power rule and the constant rule to the terms inside the parentheses:

$$ g'(t) = \sqrt t(t^0-0) $$

$$g'(t)= \sqrt t(1-0) $$ $$ g'(t)= \sqrt t$$

I am wondering if this is correct application of the the rules of differentiation. Also, I was initially tempted to apply the power rule to the $\sqrt t$ coefficient, why is it that the coefficient remains undifferentiated?

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    Is that $\sqrt{t}(t-1)$, or is it supposed to be $\sqrt{t(t-1)}$?2012-06-17
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    It is the first one2012-06-17

1 Answers 1

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When you set out to differentiate a function, look at the last operation that you would perform in computing a value of the function. In the case of $\sqrt t(t-1)$, that operation is multiplying $\sqrt t$ and $t-1$. That means that this function is (at the top level) a product and must be differentiated using the product rule:

$$g'(t)=\left[\sqrt t(t-1)\right]'=\left[\sqrt t\right][t-1]'+\left[\sqrt t\right]'(t-1)\;.$$

Now $[t-1]'=1-0=1$, exactly as you computed it in the question, but to differentiate $\sqrt t$, you’ll want to convert it to a power, $t^{1/2}$, and use the power rule:

$$\left[\sqrt t\right]'=\left[t^{1/2}\right]'=\frac12t^{-1/2}\;.$$

Now put the pieces together:

$$\begin{align*} g'(t)&=\sqrt t\cdot1+\frac12t^{-1/2}(t-1)\\ &=t^{1/2}+\frac12\left(t^{1/2}-t^{-1/2}\right)\\ &=\frac32t^{1/2}-\frac12t^{-1/2}\\ &=\frac12\left(3t^{1/2}-t^{-1/2}\right)\;. \end{align*}$$

Some instructors might prefer that you rewrite that as $$\frac12\left(3t^{1/2}-t^{-1/2}\right)=\frac12\left(3\sqrt t-\frac1{\sqrt t}\right)=\frac{3t-1}{2\sqrt t}\;.$$

Second Addition: I should have noticed that I was actually doing this the hard way. The easy way is to convert $\sqrt t$ to $t^{1/2}$ immediately and multiply out to get $g(t)=t^{1/2}(t-1)=t^{3/2}-t^{1/2}$. Now you can apply the power and sum rules to get $g'(t)=\frac32t^{1/2}-\frac12t^{-1/2}$ directly and simplify from there.

Added: If by chance your function was intended to be $g(t)=\sqrt{t(t-1)}$, the same principle applies, but the details are different. Now the last thing that you’d do in computing a value of $g$ is take the square root, so at the top level this is a power: $$g(t)=\Big(t(t-1)\Big)^{1/2}\;.$$

Thus, you’ll need to apply the power rule, combined, of course, with the chain rule:

$$g'(t)=\frac12\Big(t(t-1)\Big)^{-1/2}\Big[t(t-1)\Big]'\;.$$

You could use the power rule to do that last differentiation, but it’s much easier to multiply out and differentiate $t^2-t$ instead:

$$g'(t)=\frac12\Big(t(t-1)\Big)^{-1/2}(2t-1)=\frac{2t-1}{2\sqrt{t^2-t}}\;.$$

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    Thanks for the explanation. This question was presented in a chapter which explains the use of the constant rule, sum/difference rule, and power rule. The product rule comes up in the next chapter. I am wondering if it is possible to solve this using the former three rules, or if perhaps the product rule doesn't apply in this case?2012-06-17
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    @Kurt: Yes, it can be done very straightforwardly with just the tools that you have: multiply it out to get $t^{1/2}(t-1)=t^{3/2}-t^{1/2}$ and use the power rule on each term. Then you get $g'(t)=\frac32t^{1/2}-\frac12t^{-1/2}$ right away. I should have noticed that and said something.2012-06-17
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    Thanks again for the explanation. I am learning Calculus on my own, and at times it is difficult for me, so I am thankful for your clear explanations.2012-06-17