How do I find derivatives involving natural logarithms and multiple variables?

In summary, the conversation is about a student preparing for an exam and asking for clarification on finding the derivative of ln(x + y) and sin(x)ln(3x). They discuss using standard rules for derivatives and suggest using exponentials and implicit differentiation to solve the problems. The student also mentions confusion about using variables and constants in the problems.
  • #1
Dustobusto
32
0
So I have an exam tomorrow, and the teacher provided a review.

f(x) = ln(x + y)

I remember that

d/dx ln[f(x)] = f'(x)/f(x) so would that not equal 2/(x + y) ? The answer she gave is

1/(x + y - 1) ... where that neg. one came from I have no idea. Come to think of it, there were no problems on the homework that included two variables in this manner, so maybe the properties are slightly different?

Also, find the derivative of sin(x)ln 3x... another problem that doesn't bear much resemblance to anything on the homework, a little confused on the order of operations here.

I know that with something like 53x would be (3 ln 5)53x,

so for this one, b = sin (x). Trying to replicate it the way the book would,

(ln sin x)sin(x)ln 3x * (ln 3x)'

the derivative of ln(3x) = 3/3x, bring that to the front and get ...

I guess what I'm asking is do I use the general power rule or was I on the right track? At any point in time am I supposed to take the derivative of sin(x) and turn it into cos x?
 
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  • #2
Dustobusto said:
So I have an exam tomorrow, and the teacher provided a review.

f(x) = ln(x + y)

I remember that

d/dx ln[f(x)] = f'(x)/f(x) so would that not equal 2/(x + y) ? The answer she gave is

1/(x + y - 1) ... where that neg. one came from I have no idea. Come to think of it, there were no problems on the homework that included two variables in this manner, so maybe the properties are slightly different?

Both are incorrect and given that you have a function of x only, ie f(x) you take y as a constant (so it is still a one variable problem).
 
  • #3
I don't know where she got that -1, but I also don't know where you got the 2. y denotes a real number here, right? Not another function?

If there seems to be no way to use any of the standard rules for derivatives (the rules for sums, products, quotients, inverses and compositions), the trick is often to use the rules for exponentials to rewrite the expression you have so that you can use those rules. For example,
$$\frac{d}{dx}x^x=\frac{d}{dx}e^{\log x^x}=\cdots$$ A similar trick should work with the problem you're asking about. Another approach is to rewrite sin in terms of exponentials (and complex numbers).

I don't understand what you're saying about ##5^{3x}##.
 
Last edited:
  • #4
You can also use implicit differentiation: write the equation as ##y = sin(x)^{ln(3x)}## and take the log of both sides. Solve for ##dy/dx##.
 

What is a natural log derivative?

A natural log derivative is the derivative of a function that contains the natural log (ln) function. It is typically used to find the rate of change of a function that involves exponential growth or decay.

How do you find the derivative of ln(x)?

To find the derivative of ln(x), you can use the formula d/dx(ln(x)) = 1/x. This means that the derivative of ln(x) is equal to 1 divided by x.

What is the chain rule for natural log derivatives?

The chain rule for natural log derivatives states that if you have a function f(x) inside the natural log (ln) function, the derivative is equal to 1/f(x) multiplied by the derivative of f(x).

What is the derivative of e^x?

The derivative of e^x is e^x. This means that the rate of change of the exponential function e^x is equal to the function itself.

How do you find the derivative of a function that contains both natural log and exponential functions?

To find the derivative of a function that contains both natural log and exponential functions, you can use the chain rule. This involves taking the derivative of the outer function and then multiplying it by the derivative of the inner function.

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