Question

find f'(x) if the f(x) = the square root of 4 + e^2x

Answer 1

Ever heard of chain rule? :)

Answer 2

\[\frac{ 1 }{ 2 }\ln 4 + \frac{ 1 }{ 2}\ln e ^{2x}\]

Answer 3

\[\large f(x)=(4+e^{2x})^{\frac{1}{2}}\]
Can you use the chain rule like terenzreignz said above?

Answer 4

i thought i did.\[(4+ e ^{2x} )^{1/2} = 1/2 \ln 4 + 1/2 \ln e ^{2x} = 1/2 (4x) + 1/2(2x) = 2x + x\]

Answer 5

Any particular reason you put in a \(\ln\)?

Answer 6

what else would i use?

Answer 7

I'm pretty sure you do not need ln next to the 4.

Answer 8

Nor the e^2x as well.

Answer 9

hmm...\[\huge \frac{d}{dx}[f(x)]^n = n[f(x)]^{n-1}\cdot \frac{d}{dx}f(x)\]

Answer 10

\[(4 + e ^{2x})^{3/2} \times 2x \times 2\]

Answer 11

Are you differentiating or integrating?

Answer 12

It seems you're getting confused with that.

Answer 13

differentiating

Answer 14

You're muddled up all over the place. First off, we're differentiating. And yes, you said it yourself just now.

Answer 15

So if we're differentiating using the chain rule, we first move the exponent to the front of the expression.

Answer 16

Then we differentiate the inside. After that, we rewrite the original expression but with the exponent being less than 1 of the original exponent.

Answer 17

It's very simple if you take everything Step by STEP. I put emphasis on step by step because that's where people get stumped on. They have a need to skip steps.

Answer 18

That's where most of your silly mistakes come from. skipping steps.

Answer 19

i thought in order to bring the exponent to the front i needed a ln

Answer 20

\[1/2 (4 + e ^{2x}) = 1/2 ( 0 + e^{2x-1}) = 1/2 e^{2x-1}\]

Answer 21

Ah. I'm not sure how you thought of that, but I will help you clear everything and say that ln isn't needed in here. ;)

Answer 22

Why is e^{2x-1} in there. I feel that you though e stood for exponent.

Answer 23

The exponent is right beside the brackets/paranthesis.

Answer 24

thought*

Answer 25

and you differentiate e^2x. I think I told you this already about differentiating e. You rewrite the whole thing again and then you differentiate 2x.

Answer 26

the 2x comes from the exponent of e. Now when we use chain rule for the whole expression, you change the exponent of the whole.

Answer 27

the whole meaning the exponent right at the very end which was 1/2. What's 1/2-1?

Answer 28

the "whole", *

Answer 29

no .. im trying to work so that i use all of the rules you told me so what am i doing wrong?

Answer 30

Okay when I say differentiate the inside, what method of differentiating would you use?

Answer 31

You're using the chain rule as a WHOLE. Am I right? And in the WHOLE, you have a certain part where you differentiate the inside.

Answer 32

that certain part is where you differentiate normally.

Answer 33

That's where I think you're getting muddle up. You think when you differentiate the inside, you're also using the chain rule to differentiate.

Answer 34

Logically speaking if you were to differentiate the inside and you're differentiating e^2x, what makes yout hink you need to use the chain rule to differentiate it? hmmm...seems a bit strange, don't you think? ;)

Answer 35

you think*

Answer 36

im more confused than when i started.

Answer 37

\[\large f(x)=(4+e^{2x})^{\frac{1}{2}}\]
Okay. we have this right. the square root sign can be switched into a power/exponent which is 1/2.

Answer 38

we have this right?*

Answer 39

yes

Answer 40

Now we use the chain rule. It seems you're not really familiar with the chain rule.

Answer 41

This certain area I think you need more practice on. Not trying to lower your standards of thinking, but it is a good thing to practice on the critical parts of maths such as this.

Answer 42

Okay, the chain rule:
Let's start by moving the power/exponent to the front of the expression first.We're going step by step so you don't get confused as much.
\[\large f'(x)=\frac{1}{2}(4+e^{2x})^{\frac{1}{2}}\]

Answer 43

You confused about this first step or not?

Answer 44

I juse placed 1/2 to the front. That's your very first step.

Answer 45

if i pulled it to the front then why is it still in the exponent place?

Answer 46

I'm doing this stpe by step. How am I going to do this step by step. I have no other way because I'm on the internet. If I was teaching you in person, I would be able to teach you stpe by step. YOu just have to bare with me on this.

Answer 47

\[\large f'(x)=\frac{1}{2}(0+2e^{2x})\]
Then we differentiate the inside.

Answer 48

After that, we rewrite the original expression. THe power will become one less than before.
\[\large f'(x)=\frac{1}{2}(0+2e^{2x})(4+e^{2x})^{\frac{1}{2}-1}\]

Answer 49

Then you can get rid of the 0 and simplify that. And that will be your final answer.

Answer 50

okay .. i know what i was missing.i didn't multiply it by the original equ.

Answer 51

It's simple if you don't Over-analyse things. It's very easy to over-analyse. Okay, so you understand everything I said or are you still confused and all?

Answer 52

no i got got it.\[\frac{ e ^{2x} }{ \sqrt{4 + e ^{2x}}}\]

Answer 53

Yep. Well done. Hope you don't make the same mistake when using the chain rule again. I myself would be disappointed if I made the same mistake over and over again.

Answer 54

i get it now! thank you .. got all the other ones correct.

but how would i find the derivative of \[xe ^{y} + 1 = xy\]\[\frac{ xe ^{y} + 1 }{ x} = y\]\[xe ^{y} + 1 \times x ^{-1}\] is that correct so far?

Answer 55

differentiate with respect to x?

Answer 56

It seems that you can implicitly differentiate.
And that's not correct.

Answer 57

When you differentiate xe^y, you must use the product rule. Same goes with differentiating xy.

Answer 58

Your differentiation skills in general out a bit rusty and out of touch. But the more you practice, the better you get. The only thing I stress on would be practicing on your basics of differentiation. For example, practice on the chain rule, product rule, quotient rule, etc. All of these methods of differentiation should be considered when practicing.

Answer 59

general are*

Answer 60

first times the the dx/dy of the second plus second times the dx/dy of the first