Question

Need help to solve a quick silly question..

Answer 1

I know we use natural log...but i forgot how to apply the natural log derivative

Answer 2

I need a hint plz

Answer 3

Where did you learn that you're supposed to use natural log to find the derivative of this?

Answer 4

I remember that since the base isn’t a variable, you apply the natural log to the base and then apply the chain rule... that’s as far as i can remember

Answer 5

What you're supposed to do is let \(u = 2x + 1\). Then take the derivative of \(e^u\) with respect to \(u\) and take the derivative of \(2x + 1\) with respect to x using the chain rule. There's no natural log involved in this.

Answer 6

K so the question is to find the relative maxima and relative minima.... and i was suppose to find the first derivative of that expression and do a first derivative test to see where the graph changes from positive to negative and vice versa..

Answer 7

All you have to do is just apply the chain rule here. The only reason the derivative of \(e^x\) is \(e^x\) is because the \(x\) is treated as the "outside" function and the derivative of \(x\) is just 1. And then
\(1 \cdot e^x\) is just \(e^x\).

So with \(e^{2x+1}\) it's the same way. \(2x + 1\) is treated as the outside function and the derivative of \(2x + 1\) is just 2. So then multiply two by the inside function \(e^{2x+1}\)

Answer 8

Bro...it’s not e^2x+1 ...

Answer 9

I don't know why I saw the three as an e

Answer 10

I glanced at it too fast

Answer 11

I even said...the base is an integer... lol

Answer 12

I was already convinced it was e by the time you posted that

Answer 13

Jajajajaja

Answer 14

Shall we proceed to solve the question master? :)

Answer 15

The derivative of \(a^x\) is \(a^x \ln(a)\)

Answer 16

Wait, let me draw

Answer 17

Hang on, \(e\) IS involved in this

Answer 18

Yikes..

Answer 19

Apparently, you're supposed to use the exponent rule
\(a^b = e^{b \ln(a)}\) before applying the derivative

Answer 20

Wat? I was gonna use the log laws for some reason....you know, bring the 2x+1 to the front thingy

Answer 21

You're supposed to find the derivative of 2x + 1 when you do that

Answer 22

The answer you have is already correct. Just remove the \(x\) from the \(2x\)

Answer 23

Wait one second... let me draw

Answer 24

Like this?

Answer 25

Sorry, that was a silly mistake.

Answer 26

Correct

Answer 27

So what rule/equation do i use to find the derivative??? Plz let me know to avoid confusion

Answer 28

Showing it to you would probably confuse you

Answer 29

I guarantee it's in Larson's Calculus

Answer 30

Nope, it’s from my college textbook...

Answer 31

Literally they threw in this random expression out of nowhere... while before that i solved like 11 functions that had nothing to do with this kinda question..

Answer 32

I meant the derivative rule is in the Larson Calculus textbook.

Answer 33

Can you tell me tho :/ i need to include in my notes so i know what rule to apply the next time i study math again...next year

Answer 34

You could find it in the textbook faster than I could explain it

Answer 35

It's not exactly a simple to explain rule

Answer 36

Aww... then how will i be able to remember what to do the next time i study this course with an empty mind??

Answer 37

"next time you study this course". There's gonna be a next time? Why would you do that do yourself?

Answer 38

After that, i can just apply the power rule and collect terms

Answer 39

The DERIVATIVE of a^x is a ln(a)

Answer 40

You can't just post a^x is a^x ln(a) without proper context

Answer 41

SMH, I'm making mistakes trying to cover your mistakes

Answer 42

That was frustrating to see

Answer 43

Khekhekhekhe

Answer 44

The derivative of a^x is a^xln(a)

Answer 45

That only works if it's just \(x\) in the exponent though

Answer 46

Let me see something though.

Answer 47

K so the context is that whenever i see a base as an integer, i take the ln of the base, then apply power rule and rearrange the terms ...

Answer 48

Okay, so I did a little research and there's an alternative way to find the derivative that gives an equivalent result. (Equivalent meaning the result will look different but the expressions are still identical)

Answer 49

Here it is:
To find the derivative of \(3^{2x+1}\)
Let's say the expression is
\(a^{bx + c}\)

We re-write it as \(a^{bx} \cdot a^{c}\)

Then \(a^c\) is a constant and we can apply the constant rule \((cf)' = cf'\)

In other words
\(\boxed{(a^{bx} \cdot a^c)' = a^c(a^{bx})' = a^c [a^{bx} \ln(a) \times b]}\)

Answer 50

There's your rule

Answer 51

I see ... will i be able to apply the first derivative test on that tho ? Cuz i need to find where the expression is a relative maxima or a relative minima

Answer 52

Yes

Answer 53

I see now... k thanks mate. Really appreciated your help :)

Answer 54

Check your dm now lol