critical numbers with ln

[trigfun]

Here's the problem:

Find the critical numbers of the function.
f(x) = x^-6 ln(x)

I'm still shaky when knowing how to deal with derivatives and I'm not sure if I should use the power rule first, or use the product rule (or some version of the chain rule). My guess here is that I use the power rule first, but I wanted to see if that's the right place to start.

Thanks!

 

[Steven G]

Can we see your work for f'(x)?? We can't know where you are making a mistake if you don't show your work. After 8 posts you should know this by now.

 

[Dr.Peterson]

Here's the problem:

Find the critical numbers of the function.
f(x) = x^-6 ln(x)

I'm still shaky when knowing how to deal with derivatives and I'm not sure if I should use the power rule first, or use the product rule (or some version of the chain rule). My guess here is that I use the power rule first, but I wanted to see if that's the right place to start.

Thanks!

First, let's make sure we know what the problem says. As I read it, it's [MATH]x^{-6}\ln(x)[/MATH], but someone could think it was [MATH]x^{-6\ln(x)}[/MATH].

Now when I look at [MATH]x^{-6}\ln(x)[/MATH], I see a product of two factors, [MATH]x^{-6}[/MATH] and [MATH]\ln(x)[/MATH]. That implies the product rule is the first thing to focus on. (I'm basically looking for what the last operation is, that you would do when evaluating it -- that is, breaking it apart from the outside in.) Of course, within each part, you'll be applying whatever rule applies to that part.

 

[HallsofIvy]

You have two functions, x^6 and ln(x) multiplied together. If you were to calculate its value for a given value of x, you would calculate x^6 and ln(x) separately then multiply those values. Taking the derivative, you work "from the inside out". That is, use the product rule to get (x^6 ln(x))'= (x^6)' ln(x)+ x^6(ln(x))' then do the separate derivatives.

If this were (x ln(x))^6, so that both functions are to the 6th power, to evaluate for a given x, you would calculate x and ln(x), multiply, then take the 6th power. Again, to take the derivative, you would apply the power rule, then use the product rule.

(Notice that (x ln(x))^6 is the same as x^6(ln(x))^6 so that you could apply the product rule first to that but it would give the same result:
((x ln(x))^6)'= 6(x ln(x))^5(ln(x)+ 1)= 6x^5(ln(x))^6+ 6x^5(ln(x))^5

(x^6 (ln(x))^6)'= 6x^5(ln(x))^6+ 6x^6(ln(x))^5(1/x)= 6x^5(ln(x))^6+ 6x^5(ln(x))^5.

 

[trigfun]

First, let's make sure we know what the problem says. As I read it, it's [MATH]x^{-6}\ln(x)[/MATH], but someone could think it was [MATH]x^{-6\ln(x)}[/MATH].

Now when I look at [MATH]x^{-6}\ln(x)[/MATH], I see a product of two factors, [MATH]x^{-6}[/MATH] and [MATH]\ln(x)[/MATH]. That implies the product rule is the first thing to focus on. (I'm basically looking for what the last operation is, that you would do when evaluating it -- that is, breaking it apart from the outside in.) Of course, within each part, you'll be applying whatever rule applies to that part.

Thank you, this was helpful!

 

[trigfun]

You have two functions, x^6 and ln(x) multiplied together. If you were to calculate its value for a given value of x, you would calculate x^6 and ln(x) separately then multiply those values. Taking the derivative, you work "from the inside out". That is, use the product rule to get (x^6 ln(x))'= (x^6)' ln(x)+ x^6(ln(x))' then do the separate derivatives.

If this were (x ln(x))^6, so that both functions are to the 6th power, to evaluate for a given x, you would calculate x and ln(x), multiply, then take the 6th power. Again, to take the derivative, you would apply the power rule, then use the product rule.

(Notice that (x ln(x))^6 is the same as x^6(ln(x))^6 so that you could apply the product rule first to that but it would give the same result:
((x ln(x))^6)'= 6(x ln(x))^5(ln(x)+ 1)= 6x^5(ln(x))^6+ 6x^5(ln(x))^5

(x^6 (ln(x))^6)'= 6x^5(ln(x))^6+ 6x^6(ln(x))^5(1/x)= 6x^5(ln(x))^6+ 6x^5(ln(x))^5.

This clears things up a lot. Thank you!

 

[JeffM]

Thank you, this was helpful!

I think it greatly helps to think in terms of substitutions

[MATH]u(x) = x^{-6} \text { and } v = ln(x) \implies \\ f(x) = u(x)v(x) \implies u(x)v'(x) + u'(x)v(x).[/MATH]Now figure out u'(x) and v'(x) separately.