How can I differentiate ln equations with quotient and chain rules?

[Aerosion]

I've always had a hard time with differentiation. Always. Product rules, quotient rules, chain rules, too confusing.

Anyway, enough of my complaining. Let me post my problem.

Differentiate:

y=ln((x^3*tanx)/sq root(5x^2+3x-7))

I got some horribly long answer (which was wrong, I won't even post it, it's too absurd) and I'm wondering what I did wrong. I know I need to use the quotient rule, but how does the ln fit into that?

Help would be appreciated.

 

[Curious3141]

This is much easier broken up. Always try simplifying as much as possible before differentiating.

Try y = ln(f(x)/g(x)) = ln(f(x)) - ln(g(x))

And x^(3tanx) = e^?

And ln(\sqrt{something}) = (1/2)ln{something}

 

[FrogPad]

Go to a computer on campus and start up Maple.
Click
Tools -> Tutors -> Calculus - Single Variable -> Differentiation Methods

Now try examples in this worksheet. It will walk you through step by step how to do it. Try a lot of examples.

Trust me. Once you get the hang of it, you will see how easy it is. More than likely you are just confusing yourself, and making things harder then they really are. That's ok. It just means you don't know how to "see" the problem yet. It will just take some practice and you'll be groovy.

 

[VietDao29]

I've always had a hard time with differentiation. Always. Product rules, quotient rules, chain rules, too confusing.

Anyway, enough of my complaining. Let me post my problem.

Differentiate:

y=ln((x^3*tanx)/sq root(5x^2+3x-7))

I got some horribly long answer (which was wrong, I won't even post it, it's too absurd) and I'm wondering what I did wrong. I know I need to use the quotient rule, but how does the ln fit into that?

Help would be appreciated.

Hmm, if you encounter a product of functions, you can use Product Rule.
Example:
(x sin(x))' = x' sin(x) + x sin'(x) = sin(x) + x cos(x).
If you encounter a quotient, then you can use Quotient Rule.
Example:
\left( \frac{x}{\sin x} \right) ' = \frac{x' \sin x - x ( \sin x)'}{\sin ^ 2 x} = \frac{\sin x - x \cos x}{\sin ^ 2 x}
If you encounter something look like a composite function, then use the Chain Rule.
Example:
((x + 5)²)'.
There are many ways to do this, you can either expand all the terms out and use the Sum Rule, i.e:
((x + 5)²)' = (x² + 10x + 25)' = x²' + 10x' + 25' = 2x + 10
Or you can use the Product Rule:
((x + 5)²)' = ((x + 5) (x + 5))' = (x + 5)' (x + 5) + (x + 5) (x + 5)' = 2(x + 5).
Or you can let u = x + 5, and use the Chain Rule:
((x + 5)²)'_x = (u²)'_x = (u²)'_u u'_x = 2u(x + 5)'_x = 2u(x' + 5') = 2u = 2(x + 5).
So two ways, but they will eventually lead you to one result.
--------------
However you should consider simplifying the expression before applying the rules, or you will mess it up. By the way, is it:
y = \ln \left( \frac{x ^ 3 \tan x}{\sqrt{5x ^ 2 + 3x - 7}} \right)
or:
y = \ln \left( \frac{x ^ {3 \tan x}}{\sqrt{5x ^ 2 + 3x - 7}} \right)?
Assume that it is the first one, split it into 2, or more separate logs to get a nicer expression:
y = \ln \left( \frac{x ^ 3 \tan x}{\sqrt{5x ^ 2 + 3x - 7}} \right) = \ln (x ^ 3 \tan x) - \ln (\sqrt{5x ^ 2 + 3x - 7}) = 3\ln (x) + \ln (\tan x) - \frac{1}{2} \ln(5x ^ 2 + 3x - 7)
Now, it looks much more easier, right?
Can you go from here? :)

 

[Aerosion]

Thanks for all your help. I just have one more question.

--------------
However you should consider simplifying the expression before applying the rules, or you will mess it up. By the way, is it:
y = \ln \left( \frac{x ^ 3 \tan x}{\sqrt{5x ^ 2 + 3x - 7}} \right)
or:
y = \ln \left( \frac{x ^ {3 \tan x}}{\sqrt{5x ^ 2 + 3x - 7}} \right)?
Assume that it is the first one, split it into 2, or more separate logs to get a nicer expression:
y = \ln \left( \frac{x ^ 3 \tan x}{\sqrt{5x ^ 2 + 3x - 7}} \right) = \ln (x ^ 3 \tan x) - \ln (\sqrt{5x ^ 2 + 3x - 7}) = 3\ln (x) + \ln (\tan x) - \frac{1}{2} \ln(5x ^ 2 + 3x - 7)
Now, it looks much more easier, right?
Can you go from here? :)

So I should just apply one of the rules to that simplified equation, or should I just solve it directly?

If it's the latter, I could definitely do that, but if you have to apply a rule, then what would that be? The whole simplified problem only adds and subtracts the ln equations, and the rules specify only quotients, products, or powers, so how could any rule even be used here?

 

[VietDao29]

Thanks for all your help. I just have one more question.



So I should just apply one of the rules to that simplified equation, or should I just solve it directly?

If it's the latter, I could definitely do that, but if you have to apply a rule, then what would that be? The whole simplified problem only adds and subtracts the ln equations, and the rules specify only quotients, products, or powers, so how could any rule even be used here?

Ok, the 2 expressions are the same, so their derivatives will be the same. So will you choose the easier one to evaluate or you want to do the harder one? Of course, the easier one, right? :)
Yes, you can apply the 3 rules directly on the original expression, but the result will turn out to be very ugly, and you will have to simplify that, and you may mess it up. So why don't consider simplifying the expression before differentiate it?
By the way, you didn't answer my question, is it:
y = \ln \left( \frac{x ^ 3 \tan x}{\sqrt{5x ^ 2 + 3x - 7}} \right) or y = \ln \left( \frac{x ^ {3 \tan x}}{\sqrt{5x ^ 2 + 3x - 7}} \right)?
And you may need to use the Chain Rule to take the derivative of \frac{1}{2} \ln(5x ^ 2 + 3x - 7).
Can you go from here? :)