How to find derivative in radical function?

Question

I need to find derivatives of following functions:

$$\frac32x^\frac32-\frac{2x^2}{3}$$

$$-2\sqrt{x}-\frac{-2}{\sqrt{x}}$$

So starting from first one, I have tried to first simplify the fractions to have same denominator:

$$\frac96x^\frac32-\frac{4x^2}{6}$$

Then derive using the power rule:

$$\frac96 \cdot \frac32x^\frac12-\frac{2 \cdot 4x}{6}$$

It results in something like this:

$$\frac{27}{12}x^\frac12-\frac{8x}{6}$$

Multiply the right side by two to get same denominator:

$$\frac{27}{12}x^\frac12-\frac{16x}{12}$$

Simplify:

$$27x^\frac12-16x$$

Is this the correct derivative for first function? I have no clue. Second function, here I have tried to use formula:

$$\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$$

So first find derivatives for $f(x)$ and $g(x)$ \begin{align*} f & = -2\sqrt{x}-2\\ f' & = \frac{-2}{2\sqrt{x}}\\ g & = \sqrt{x}\\ g' & = \frac{1}{2\sqrt{x}} \end{align*}

Then construct the formula:

$$\frac{\frac{-2}{2\sqrt{x}} \cdot \sqrt{x}-(-2\sqrt{x}-2) \cdot \frac{1}{2\sqrt{x}}}{\sqrt{x}^2}$$

Unfortunately I was not able to take this any further. Could I have some advise please?

Answer 1

Some tips:

1. You can use the power rule even though your fractions don't have the same denominator $$\frac32x^\frac32-\frac{2x^2}{3}$$ You can apply the power rule directly to the expression above to get $$ \frac32*\frac32x^\frac12-2*\frac{2x}{3}$$ Which gives you $$ \frac94x^\frac12-\frac{4x}{3}$$ This is the correct answer for the first part.

2. You can convert $$-2\sqrt{x} - \frac{-2}{\sqrt{x}}$$ to its power form (similar to part 1)which is $$-2x^\frac12-(-2)x^\frac{-1}{2}$$ and then apply the power rule to derive it.

3. I would suggest you to write your work in equation form rather than expression form as you seem to have confused the two. For example, in your work

Multiplay the right side by two to get same denominator: $$\frac{27}{12}x^\frac12-\frac{16x}{12}$$

Simplify:

$$27x^\frac12-16x$$

You are totally correct until you got to the simplify part. We DO NOT remove the denominator of an EXPRESSION by multiplying it.

You may have gotten this habit from doing equations where this method is correct. As illustrated

$$y=\frac{27}{12}x^\frac12-\frac{16x}{12}$$ $$12y=27x^\frac12-16x$$

But when you're doing calculations in expression form, this habit is very dangerous.

And as we all know y is not equal to 12y, thus your answer being incorrect.

Anyway the correct way of doing simplification is dividing both the numerator and the denominator by their common multiple which in the case of 27/12 is 3 $$\frac{\frac{27}3}{\frac{12}3} = \frac94$$ Which implies that $$\frac{27}{12}x^\frac12-\frac{16x}{12} = \frac{9}{4}x^\frac12-\frac{4x}{3}$$

Answer 2

The derivative of $x^{\alpha}$ is $\alpha x^{\alpha-1}$, for al $\alpha \neq 0$.

Answer 3

The derivative of $$ \frac32x^{3/2}-\frac{2x^2}{3} $$ is $$ \frac94x^{1/2}-\frac{4x}{3} . $$ Your steps were OK until you said "simplify"; then you lost your denominator for no reason. In doing problems like this there is no need for common denominator. Just apply the power rule for each term separately.