Question

find the derivative of (3x^2)+ (2/(3x^2))

Answer 1

multiply by the power, then take 1 from the power, if just an integer, then it is removed.

there if y = 3 + 2/3x^2

then dy/dx = 4/3x

Answer 2

for some reason I had 3-(4/3x^2)

Answer 3

the question was 3+(2/3x^2)

Answer 4

Even that isn't completely unambiguous. Do you mean
\[3+\frac{2}{3x^2}\]or\[3+\frac{2}{3}x^2\]Use the equation editor or enough parentheses to make it unambiguous. If you meant the latter, writing 3+2x^2/3 is perhaps clearer (then the left to right evaluation gets you the right thing, so long as you didn't mean to raise \(x^{2/3}\)! No question at all about the typeset versions, though.

\[\frac{d}{dx}[3+\frac{2}{3x^2}] = -\frac{4}{3x^3}\]
\[\frac{d}{dx}[3+\frac{2}3x^2]=\frac{4}{3}x\]

Answer 5

it's the first
\[3x^2+\frac{ 2 }{ 3x^2 } \]

Answer 6

I don't see that anywhere else :-)

\[\frac{d}{dx}[3x^2+\frac{2}{3x^2}] = 6x -\frac{4}{3x^3}\]

Answer 7

@whpalmer, could you please show me the steps.

Answer 8

Sure, I'll use the last one I did to demonstrate:

\[\frac{d}{dx}[3x^2+\frac{2}{3x^2}]\]First, I'll rewrite the second term with a negative exponent\]
\[\frac{d}{dx}[3x^2+\frac{2}{3}x^{-2}]\]Now, the rule for differentiating \(x^n\) is simply \[\frac{d}{dx}[x^n] = nx^{n-1}\]so marching through the terms, we get
\[\frac{d}{dx}[3x^2+\frac{2}{3}x^{-2}= 3*2x^{2-1} + (-2)\frac{2}{3}x^{-2-1}=6x-\frac{4}{3}x^{-3}=6x-\frac{4}{3x^3} \]

Answer 9

nuts, I missed the closing bracket on the left side of the last line, but I trust you can imagine it there :-)