derivative help please!?

log_9 means "logarithm to the base 9." For example, log9( 9 ) = 1, which means that 9¹ = 9. ("9 to the power 1 is 9.") You can do logarithms to any base you want. It asks the question, "If B and x are given, what is n if Bⁿ = x?" So-called "natural logarithms" use base B=e and the "common logarithm" uses B=10. You've been thrown a curve ball, asking, "Given x, what is n for 9ⁿ = x?" Your teacher imagines you know all this stuff. So you need to know it.

Let's start with this:

1.    y = bª

You should have some idea about raising one number to a power. If not, you need to grasp that before continuing.

Then remember this one:

2.    ln y = a ln b

Now, I'm using ln to suggest the natural logarithm. But I could just as well have used the common logarithm (which I usually call 'log' when the context is understood), or even your "base 9" logarithm. It's all the same thing. It doesn't matter which logarithm you use, so long as you use the _same_ one for both sides, you are okay.

There is a reason for (2). Multiplying two numbers is the same as adding up their logarithms and then using that sum as the power for the base used in the logarithms. So when you raise something to the power of 2, for example, that is the same as multiplying the number by itself, which is the same as adding up the logarithm twice, or multiplying it by 2. So the power multiplies the logarithm. But just remember it, if nothing else.

To help you remember, I'll add this note. This knowledge was very practical "back in the day" before calculators. Folks would look up the logarithms of some two numbers they were multiplying, in a table, then add those two numbers, then go back into a table and look up sum to find the product. It was relatively easy to add lots of numbers rather than multiplying them. But today, all that is a bit of a lost art.

Okay. So let's continue. Remember that I said it doesn't matter which logarithm you use, except that you use the same one on both sides? So let's look at (2), slightly differently. Let's introduce a new notation, too -- let's say that log_b means "logarithm to the base b":

3.    a = ln x / ln b = log x / log b = log9 x / log9 b = log_b x

or, in short:

4.    log_b x = ln x / ln b = log x / log b = log9 x / log9 b = ...

So, here you can say that given value x and base b, you can compute a using any one of many different kinds of logarithms. That's kind of neat. It means that if your calculator only does common logarithms, you can still compute the natural logarithm. Or if your calculator doesn't do "log to the base 9" then you can still compute that with whatever logarithm you DO happen to have.

So (4) above is VERY USEFUL to know about.

By the way, note the following. If log_b means "logarithm to the base b" then:

5.    b^(log_b y) = b^(a log_b b) = b^(a) = bª = y

Okay. These are basics you are supposed to know _before_ you learn calculus. Not because that is the best time, so much as that teachers of calculus generally assume you've got all this stuff down by the time you reach their class. They can be wrong about that and you can still do a lot in calculus without knowing all this. But it helps to know it, because the teachers will generally not cover "old ground." Most of their students will have already accumulated all this beforehand.

So rephrase your function:

6.    f(x) = x² log_9( x² + 9 ) = x² [ ln( x² + 9 ) / ln 9 ]

Do you see? We just used (4) above so that we can go back to using the natural logarithm, where the derivative is usually stated explicitly in books and tables and so on. Since (1/ln 9) is a constant, the only thing it does to you is to place a constant times your resulting derivative without it. So it's really easy to deal with. Like this:

7.    d( f(x) ) = d{ x² [ ln( x² + 9 ) / ln 9 ] } = (1/ln 9) • d{ x² [ ln( x² + 9 ) ] }

8.    d( f(x) ) = (1/ln 9) • [ d{ x² } ln( x² + 9 ) + x² d{ ln( x² + 9 ) } ]

9.    d( f(x) ) = (1/ln 9) • [ { 2x dx } ln( x² + 9 ) + x² { 2x dx / ( x² + 9 ) } ]

10.   d( f(x) ) = (1/ln 9) • [ 2x ln( x² + 9 ) + 2x³ / ( x² + 9 ) ] dx

11.   d( f(x) ) = (1/ln 9) • 2x • [ ln( x² + 9 ) + x² / ( x² + 9 ) ] dx

12.   d( f(x) ) / dx = (2x/ln 9) • [ ln( x² + 9 ) + x² / ( x² + 9 ) ]

I hope all that makes sense and that the above lesson allows you to do your own work on problems like this.

You could distribute (1/ln 9) back across to get:

13.   d( f(x) ) / dx = 2x • { ln( x² + 9 ) / ln 9 + x² / [ ( x² + 9 ) • ln 9 ] }

14.   d( f(x) ) / dx = 2x • { log_9( x² + 9 ) + x² / [ ( x² + 9 ) • ln 9 ] }

But it's better to use natural logarithms and pull the constant out front. Or else you have to keep track of small details as things go to say the 2nd derivative, etc. Best to do the conversion once, haul the constant out front, and _never_ deal with it after.

Best wishes.

Product rule:

f'(x) = x^2 * (2x) / ( (x^2+9)*ln(9) ) + log9(x^2+9) * 2x

(the other answers forgot the ln(9) )

Recall: the derivative of log_a(x) = 1 / ( x * ln(a) )

Use the property of logarithm as Loga(x)=Loga(b)*Logb(x)

As, log9(x^2 + 9)= log9(e) loge(x^2 + 9)

So the expression becomes, f(x)=x^2 log9(e) loge(x^2 + 9)

Using product rule,

df(x)/dx=log9(e) {loge(x^2+9) dx^2/dx + x^2 dloge(x^2+9)/dx}

=log9(e) {loge(x^2+9) 2x + x^2 dloge(x^2+9)/d(x^2+9) * d(x^2+9)/dx} using chain rule

=log9(e) {2x loge(x^2+9) + x^2 *1/(x^2+9) *(2x)}

Simplify the terms

For any derivative, you need to convert to the natural log (ln). To covert log base 9(log_9), use the identity, log_b(a) = {1/ln(b)}ln(a). f(x) becomes:

f(x) = {1/ln(9)}x²{ln(x² + 9)}

Notice that 1/ln(9) is just a constant and the rest can be done by the product rule and the chain rule.

First the product rule:

{g(x)h(x)}' = g'(x)h(x) + g(x)h'(x)

let g(x) = {1/ln(9)}x², then g'(x) = {2/ln(9)}x

let h(x) = ln(x² + 9), then h'(x) must be done using the chain rule:

h'(u(x)) = dh/du•du/dx

let u = x² + 9, then du/dx = 2x

dh/du = d[ln(u)]/du = 1/u = 1/(x² + 9)

h'(u(x)) = 2x/(x² + 9)

{g(x)h(x)}' = {2/ln(9)}x{ln(x² + 9)} + {1/ln(9)}x²{2x/(x² + 9)}

f'(x) = {2x/ln(9)}{ln(x² + 9)} + x²/(x² + 9)}

That's a base 9 log.

You can use change of base formula before differentiating:

log_base_b (a) = log(a)/log(b) or ln(a)/ln(b) or log_base_c (a) / log_base_c (b)

f(x) = x² log₉(x²+9)

f(x) = x² ln(x²+9)/ln(9)

f(x) = 1/ln(9) x² ln(x²+9)

Now 1/ln(9) is just a constant, so f'(x) becomes:

f'(x) = 1/ln(9) [d/dx (x²) * ln(x²+9) + x² * d/dx (ln(x²+9))]

f'(x) = 1/ln(9) [2x ln(x²+9) + x² * 1/(x²+9) * 2x]

f'(x) = 1/ln(9) [2x ln(x²+9) + 2x³/(x²+9)]

Mαthmφm

let log be log to base 9

f (x) = x² log (x² + 9)

f ` (x) = (2x) log (x² + 9) + [ 2x/(x² + 9) ] (2x)

f ` (x) = (2x) log (x² + 9) + 4x² / (x² + 9)