Thread: determining the derivative of exponental functions

how do you find the derivative of these exponential functions? also what is "1n" exactly? is it equal to "e"?which has a value of 2.718.



Thx in advance to anyone who helps.

Hello, math619!

You're scary . . .

First of all, these are not exponential functions.

Second, it's not $\displaystyle 1n$
It's $\displaystyle ln\\$ which stands for "natural log".

And you're expected to know the derivative of a log function.
(Knowing Log Properties and the Chain Rule couldn't hurt.)

It sounds like you missed a few classes.

Originally Posted by math619

how do you find the derivative of these exponential functions? also what is "1n" exactly? is it equal to "e"?which has a value of 2.718.



Thx in advance to anyone who helps.

As Soroban says its $\displaystyle \ln$ not $\displaystyle 1n$. That is its the natural logarithm, or log
to the base $\displaystyle e$. That means if:

$\displaystyle y=\ln(x)$,

then:

$\displaystyle e^y=x$

Then the property that you need to find the derivatives of your functions is (you will also need the chain rule):

$\displaystyle \frac{d}{dx}\ln(x)=\frac{1}{x}$

RonL

Originally Posted by Soroban

Hello, math619!

You're scary . . .

First of all, these are not exponential functions.

Second, it's not $\displaystyle 1n$
It's $\displaystyle ln\\$ which stands for "natural log".

And you're expected to know the derivative of a log function.
(Knowing Log Properties and the Chain Rule couldn't hurt.)

It sounds like you missed a few classes.

Lol, you make me chuckle Soroban.

Any way, lets see what we can do.

Originally Posted by math619

how do you find the derivative of these exponential functions? also what is "1n" exactly? is it equal to "e"?which has a value of 2.718.



Thx in advance to anyone who helps.

But yes, 1n is definitely a first for me.

a.) y = ln(x^2 + 3x + 4)

The derivative of ln(x) is 1/x.

But, we have a composition of functions. Thus, we need to take the derivative and multiply it, too, of the inside.

Thus, 1/(x^2 + 3x + 4) * 2x + 3 = (2x + 3)/(x^2 + 3x + 4)

b.) You need to clarify this one:

y = log(x^2) or y = log(x)^2?

Also, log base 10 I am assuming?

c.) y = ln((x^2)/(3x - 4)^3)

1/((x^2)/(3x-4)^3) * derivative of inside

= ((3x - 4)^3)/(x^2) * quotient rule

Recall the quotient rule: (f(x)/g(x))' = (g(x)*f'(x) - f(x)*g'(x))/(g(x)^2)

((3x-4)^3*2x - [x^2*3(3x-4)^2*3])/[((3x-4)^3)^2]

I'll let you simplify: (-3x^2 - 8x)/(3x - 4)^4

Thus,

((3x - 4)^3)/(x^2)*(-3x^2 - 8x)/(3x - 4)^4

Simplified: (-3x - 8)/(3x^2 - 4x)

Or you can expand it: 2/x - 9/(3x - 4)

Both are equivalent.