# Differentiating Logarithmic and Exponential Functions

1. Jun 1, 2006

### nothing123

a question i couldnt get the right answer to:

Differentiate
1. y= 3^x (log(3) x)

[3 = base (dont know the right syntax for typing up logs on the web)]

what i have so far:
y = 3^x (ln 3) (log(3) x) + 3^x / xln3

right answer: 3^x (ln 3) + 3^x / xln3

if we apply the product rule, where did the (log(3) x) go?

2. Jun 1, 2006

### dav2008

Are you sure the right answer isn't $$(3^x)ln(x)+ \frac{3^x}{x\cdot ln(3)}$$?

In that case it's just a simplified form of your answer where you write $$log_3(x)$$ as $$\frac{ln(x)}{ln(3)}$$ and the ln(3) cancels out.

Last edited: Jun 1, 2006
3. Jun 1, 2006

### nothing123

yes, it is...you're absolutely right. thanks!

4. Jun 2, 2006

### nothing123

quick question again, how does 2/ln4 = 1/ln2?

5. Jun 2, 2006

### dav2008

Well think about it. Can you use any of the properties of logs to write ln(4) in terms of ln(2)?

6. Jun 2, 2006

### nothing123

ahhh...ln(4) = ln(2)^2 = 2ln(2).

i have one more problem im having trouble differentiating

y = x(3x)^(x^2) - i hope thats understandable (basically 3x to the power of x^2)

so y' = (3x)^(x^2) + x(3x)^x^2*ln(3x)* 2x*3

i just did product rule and chain rule but this is not even close to the right answer which is: x(3x)^x^2 * [1/x + x + 2x*ln(3x)]

i dont know how they got 3 terms inside the square brackets when theres only one set of product rule (i.e f'g + fg')

7. Jun 2, 2006

### nothing123

sorry one more:

Differentiate
y = x^(1/ln(x))

y' = 1/ln(x) * x^(1/ln(x) - 1)

just doing power rule here but the answer is 0

8. Jun 2, 2006

### Pseudo Statistic

y = x^(1/ln(x))...
Apply ln to both sides...
ln y = 1 / ln x * ln x
ln y = 1
y'/y = 0 (Chain rule)
y' = 0.

Last edited: Jun 2, 2006
9. Jun 2, 2006

### TD

The fact that you find 0 means that y had to be a constant, right?
Now if y = x^(1/ln(x)) then ln(y) = 1/ln(x).ln(x) = 1, as shown above.
Well, if ln(y) = 1, then y has to be e so x^(1/ln(x)) is actually e

10. Jun 2, 2006

### VietDao29

There's one more way, which can be applied to this problem. There's a property of logs faying:
$$\log_a b = \frac{1}{\log_b a}$$
This can be proved by using the fact that 1 = logbb
$$\Leftrightarrow \log_a b = \frac{\log_b b}{\log_b a}$$
$$\Leftrightarrow \log_a b = \log_a b$$ (which is true)
So applying that property here, we have:
$$x ^ {\frac{1}{\ln x}} = x ^ {\log_x e} = e$$
Now, can you differentiate it? :)

11. Jun 2, 2006

### Pseudo Statistic

My way's the coolest. ;)