Question

Check my work for derivative

Answer 1

I dont have to simply further, but is the math right?

Answer 2

Nope.$$\frac{\log x}{x^3}\ne\log\frac{x^3}x$$

Answer 3

better pic

Answer 4

My point still stands.

Answer 5

\[\huge y= \frac{ lnx }{ x^3 }= lnx-linx^3\]

Answer 6

What if i take the ln of both sides?

Answer 7

That's not a valid identity.$$\log\frac{x}{x^3}=\log x-\log x^3$$... but \(\dfrac{\log x}{x^3}\ne\log\frac{x}{x^3}\)

Answer 8

Right, i understand what you're saying now.

Answer 9

So i would have to use the quotient rule?

Answer 10

No point. Just use the product rule:$$y=\frac1{x^3}\log x=x^{-3}\log x\\\frac{dy}{dx}=\left(\frac{d}{dx}x^{-3}\right)\log x+x^{-3}\left(\frac{d}{dx}\log x\right)=-3x^{-4}\log x+x^{-3}\cdot\frac1x$$

Answer 11

... which simplifies nicely: $$-\frac3{x^4}\log x+\frac1{x^4}=\frac{1-3\log x}{x^4}$$

Answer 12

But would the quotient rule also work?

Answer 13

The quotient rule will work but it's a lot more pain to use than the product rule so there's no reason to prefer it :-p

Answer 14

Yeah you're are right , but the thing is my teacher wants us to practice it more :P

Answer 15

ok$$y=\frac{\log x}{x^3}\\\frac{dy}{dx}=\frac{(\frac{d}{dx}\log x)x^3-\log x(\frac{d}{dx}x^3)}{(x^3)^2}=\frac{x^2-3x^2\log x}{x^6}=\frac{x^2(1-3\log x)}{x^6}=\frac{1-3\log x}{x^4}$$

Answer 16

Thanks I got it both ways !!

Answer 17

Is this done right?

Answer 18

@oldrin.bataku

Answer 19

hm$$y=(\underbrace{x+\log x}_u)^3=u^3\\\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=3u^2\cdot\left(1+\frac1x\right)=3(x+\log x)^2\left(1+\frac1x\right)$$

Answer 20

Close but your derivative for \(u\) was incorrect:$$\frac{du}{dx}=\frac{d}{dx}[x]+\frac{d}{dx}[\log x]=1+\frac1x$$