Differentiating Logarithmic and Exponential Functions

In summary, the conversation is about differentiating equations and finding the correct answer. The first problem involves differentiating y=3^x(log(3)x), with the correct answer being 3^x(ln3)+3^x/xln3. The second problem involves differentiating y=x(3x)^(x^2), with the correct answer being x(3x)^x^2 * [1/x + x + 2x*ln(3x)]. The third problem involves differentiating y=x^(1/ln(x)), with the correct answer being 0. The conversation also discusses the use of properties of logs and the application of the chain rule.
  • #1
nothing123
97
0
a question i couldn't 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?
 
Physics news on Phys.org
  • #2
Are you sure the right answer isn't [tex](3^x)ln(x)+ \frac{3^x}{x\cdot ln(3)}[/tex]?

In that case it's just a simplified form of your answer where you write [tex]log_3(x)[/tex] as [tex] \frac{ln(x)}{ln(3)}[/tex] and the ln(3) cancels out.
 
Last edited:
  • #3
yes, it is...you're absolutely right. thanks!
 
  • #4
quick question again, how does 2/ln4 = 1/ln2?
 
  • #5
Well think about it. Can you use any of the properties of logs to write ln(4) in terms of ln(2)?
 
  • #6
ahhh...ln(4) = ln(2)^2 = 2ln(2).

i have one more problem I am having trouble differentiating

y = x(3x)^(x^2) - i hope that's 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 don't know how they got 3 terms inside the square brackets when there's only one set of product rule (i.e f'g + fg')
 
  • #7
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
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:
  • #9
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 :smile:
 
  • #10
nothing123 said:
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
There's one more way, which can be applied to this problem. There's a property of logs faying:
[tex]\log_a b = \frac{1}{\log_b a}[/tex]
This can be proved by using the fact that 1 = logbb
[tex]\Leftrightarrow \log_a b = \frac{\log_b b}{\log_b a}[/tex]
[tex]\Leftrightarrow \log_a b = \log_a b[/tex] (which is true)
So applying that property here, we have:
[tex]x ^ {\frac{1}{\ln x}} = x ^ {\log_x e} = e[/tex]
Now, can you differentiate it? :)
 
  • #11
My way's the coolest. ;)
 

What is the difference between logarithmic and exponential functions?

Logarithmic functions involve taking the logarithm of a number, while exponential functions involve raising a number to a power.

What is the general form of a logarithmic function?

The general form of a logarithmic function is y = logb(x), where b is the base of the logarithm and x is the input value.

What is the general form of an exponential function?

The general form of an exponential function is y = bx, where b is the base of the exponential and x is the input value.

How can I tell if a function is logarithmic or exponential?

If the variable is in the exponent, then the function is exponential. If the variable is in the base, then the function is logarithmic.

What is the relationship between logarithmic and exponential functions?

Logarithmic and exponential functions are inverse functions of each other. This means that if you apply a logarithm to an exponential function, you will get the original input value, and vice versa. In other words, they "undo" each other.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
191
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
780
  • Calculus and Beyond Homework Help
Replies
24
Views
2K
Replies
4
Views
695
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
170
  • Calculus and Beyond Homework Help
Replies
7
Views
640
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top