Finding f'(x) of f(x) = (x^2-3)/(7ln[x]+3)

[Twin468]

Homework Statement

f(x) = (x^2-3)/(7ln[x]+3) find f'(x)

Homework Equations

The Attempt at a Solution

My attempt :
(7ln[x]+3)(x^2-3)' - (x^2-3)((7ln[x]+3)' / (7ln[x]+3)^2

(7ln[x]+3)(2x) - (x^2-3)((7/x) / (7ln[x]+3)^2

(14xln[x]+6x) - ((7x^2-21)/x) / (7ln[x]+3)^2


Is this right??

I am so lost...

 

[Mentallic]

Yeah it's right. Why do you feel so lost?

 

[Twin468]

Thanks for the quick reply. Trying to study for finals but this did not feel like an adequate answer.
Glad to hear I am on the right track.

 

[Twin468]

Do you have time for one more?

 

[Twin468]

f(x) = ln(5x^3 - 5)^3 find f'(x)

My solution :
u = (5x^3 - 5) so du is 15x^2
ln(u)^3(du)

3ln(5x^3 - 5)^2(15x^2) ??

Is that right??

 

[Mentallic]

No worries. Good luck on the exams!

 

[Mentallic]

Oh that depends, is it ln\left((5x^3-5)^3\right) or \left(ln(5x^3-5)\right)^3.

From what you've done, it looks like the latter.

 

[Twin468]

it is y = ln(5x^3-5)^3
so then the answer I had is right?

 

[Gib Z]

f(x) = ln(5x^3 - 5)^3 find f'(x)

My solution :
u = (5x^3 - 5) so du is 15x^2
ln(u)^3(du)

3ln(5x^3 - 5)^2(15x^2) ??

Is that right??

That's not right. The derivative of ln(u)^3 is not 3ln(u)^2. Also, could you clarify whether you mean f(x) = \ln ( (5x^3-5)^3) or f(x) = ( ln (5x^3-5) )^3.

 

[Twin468]

It is the second one you have posted... f(x) = (ln(5x^3-5))^3
(how do you get the superscript?)

 

[Gib Z]

Do you know how to simplify ln (a^b)? That will help you for this question.

 

[Mentallic]

Gib Z, the question is exponentiating the entire logarithm.

For y=\left(ln(5x^3-5)\right)^3

Let u=5x^3-5

Now we have y=\left(ln(u)\right)^3

Let t=ln(u)

So finally we have y=t^3

Now the derivative is \frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{du}.\frac{du}{dx}

 

[Twin468]

Would it be bln(a) ?

 

[Mentallic]

Yes ln(a^b)=b.ln(a) but that's not the problem at hand, according to:

It is the second one you have posted... f(x) = (ln(5x^3-5))^3
(how do you get the superscript?)

 

[Twin468]

Ok, I will use your previous post to figure it out. Thanks Mentallic! I really appreciate it!

 

[Mentallic]

No problem

By the way, I never use this long tedious method of substitution to find the derivative. Finding the pattern in what happens will really save you time and energy.

Let's look at your most recent example:

y=\left(ln(5x^3-5)\right)^3

First we picture the entire problem as (something)³, and so we use the usual rule and arrive at:

\frac{dy}{dx}=3\left(ln(5x^3-5)\right)^2 --- not the answer

but now we need to multiply everything by the derivative of what's inside, and then what's inside that... etc. till we reach a finishing point.

we look at what's inside as being ln(something), so we have to multiply by the derivative of that, which, by our rule with logarithms becomes 1/(something).

Now we have:

\frac{dy}{dx}=3\left(ln(5x^3-5)\right)^2.\frac{1}{5x^3-5} --- not the answer

But again, we've ignored what's inside of that log, so we need to multiply by the derivative of that as well,

So finally we end up with:

\frac{dy}{dx}=3\left(ln(5x^3-5)\right)^2.\frac{1}{5x^3-5}.(15x^2) --- the answer

We can even go again and think of it as 5(something)³-5 but then the derivative of that something is 1 so it doesn't change anything. This is our true finishing point since there are no more variables to deal with.