Question

fid the derivative:

y=e^sin^2(lnx)

Answer 1

\[y = e^{\sin^{2} (lnx)}\]?

Answer 2

yes

Answer 3

Let's use the chain rule"
\[y = e^{u} \frac{du}{dx}\]

Answer 4

dy/dx = ...*

Answer 5

So yeah, let's find the derivative of the power, first.

Answer 6

\[\sin^{2}(lnx)\] Some more chain rule here.

Answer 7

u= lnx ?

Answer 8

Uh, yeah, for this part.

Answer 9

u= 1/x

Answer 10

\[e^{\sin ^{2}\ln x} 2\sin( \ln x) \cos (\ln x)(1/x)\]

Answer 11

Is this the right solution?

Answer 12

I believe so. You see, we kept the e^u part, but we multiply it by the derivative of u to get the derivative of the entire thing.

Answer 13

soso707 then what is the right solution?????

Answer 14

Many chain rules.

Answer 15

but I don't understand? can you understand me >>?

Answer 16

I want steps for this Question

Answer 17

Ok. Let's go back to the general form:
\[\frac{dy}{dx} = e^{u} \frac{du}{dx}\]

Answer 18

As you see, the derivative of an exponential function with e involves us keep the original function, but multiplying is by the derivative of its power.

Answer 19

So, we're doing this:
\[\frac{dy}{dx} = e^{\sin^{2}(lnx)}* \frac{d(\sin^2 (lnx))}{dx}\]

Answer 20

ok, I think I'm understand

Answer 21

Great. Can you find the derivative of the second part, now?

Answer 22

which second part?

Answer 23

\[\frac{dy}{dx} = e^{\sin^{2}(lnx)}* \frac{d(\sin^2 (lnx))}{dx}\]

Answer 24

The "second" or "latter" part is \[ \frac{d(\sin^2 (lnx))}{dx}\]

Answer 25

Please try that. Remember the power rule? Use that first.

Answer 26

(sin^2 . 1/x) + (2cos . lnx)

?

Answer 27

would you mind doing this easier problem first?\[\frac{ d }{ dx }(2\cos ^{2}x)\]

Answer 28

I'm asking you to do this for practice in using the power and chain rules.

Answer 29

-4sinx

Answer 30

Here's how I'd approach that example problem:

1. Take the coefficient (2) out:
\[2\frac{ d }{ dx }(\cos ^{2}x)\]

Answer 31

ok ..

Answer 32

2. Think of \[\cos ^{2}x. as. (\cos x)^2\]

Answer 33

Find the derivative of \[(\cos x)^{2}\]

Answer 34

Remember to use the power rule and chain rules here.

Answer 35

2(cosx) -sinx
if you write like this, that true ?

Answer 36

You're on the right track. There's a set of parentheses missing in your 2(cosx) -sinx. Can you figure out where the parentheses are missing and insert them, please?

The derivative of (cos x)^2 is :

Answer 37

2(cosx)(-sinx)

like that ?

Answer 38

so much better. That's perfect. Without the 2nd set of parentheses, it'd appear that you were subtracting sin x instead of multiplying by (-sin x).

OK. Now let me change the problem a bit by replacing x with a function:

find the derivative of (cos ln x)^2.
Remember to use the power rule first, and then the Chain Rule, and to use those lovely parentheses!

Answer 39

2(cos ln x) (cos . 1/x + (-sin . ln x))

Answer 40

\[\frac{ d }{ dx }(\cos \ln x)^{2}=2(\cos \ln x)^{1}\frac{ d }{ dx }\ln x\]

Answer 41

First you take the derivative of (the power function (cos ln x)^2, which you have done correctly; the correct result is 2(cos ln x). But after that you must apply the chain rule and find the derivative of ln x:

Answer 42

2(cos lnx)^1 (1/x) ?

Answer 43

Really nice. You're right on target. What have you learned here?

Answer 44

Err, what about the -sin(lnx)

Answer 45

Err, I overlooked that, and you've overlooked the necessary parentheses again. Mind going back and fixing up your result to include the derivative of cos ln x?

Answer 46

I learned 2 rules

Answer 47

Your initial response was "2(cos ln x) (cos . 1/x + (-sin . ln x))" Can you fix that?

Answer 48

First you'll want to write that = (cos ln x), followed by the derivative of cos ln x, followed by the derivative of ln x. Try that , please.

Answer 49

2(cos ln x) (1/x) ?

Answer 50

See my previous comment.

Answer 51

As a review, here's what we have so far in this EXAMPLE problem:\[\frac{ d }{ dx }(\cos \ln x)^2=2(\cos \ln x)^1( ? )( ? ))\]

Answer 52

2(cos ln x)(cos lnx)(1/x)

Answer 53

Your 2(cos ln x)(cos lnx)(1/x) is on the right track, but you haven't yet found the derivative of cos ln x, have you? thus, throw out your (cos lnx) term (in the middle) and replace that with the derivative of (cos ln x).

Answer 54

can you the true sulotion for this 2(cos ln x) (cos . 1/x + (-sin . ln x)) ?

Answer 55

write *

Answer 56

I'll take what you have: 2(cos ln x)(cos lnx)(1/x)

and write out a corrected version: 2(cos ln x)(-sin lnx)(1/x). That's it.
Note that this problem we've been working on is an EXAMPLE; we have not found the derivative of the original problem.

Answer 57

Original problem:

Find the derivative of y=e^sin^2(lnx), or \[y = e^{\sin^{2} (lnx)}\]

Answer 58

First, what is the derivative of e^x?

Answer 59

xe

Answer 60

Hi, Winner66! Of course. Go right ahead!

Answer 61

@soso707: Would you please look up "derivative of exponential functions" .
Winner66 is presenting an alternative way of finding the derivative, a good and valid method. However, I'd like to finish the original approach before considering an alternative approach.

What is the derivative of e^x?

Answer 62

@loser66: Your method will, of course, succeed, but I think it would be helpful to @soso707 if he and I were to finish applying the original approach. Once that's clear, we could consider your suggested "logarithmic differentiation." Thanks.

Answer 63

First, what is the derivative of e^x? Answer: e^x.
Second, what is the derivative of e^(2x)? @soso707 ?

Answer 64

yes thanks :)

Answer 65

Second, what is the derivative of e^(2x)? @soso707 ?

Answer 66

We need the Chain Rule here.

the derivative of e^x is simply e^x.

The derivative of e^u, where u is a separate function of x, is e^u*(du/dx). Have you seen this before?

Answer 67

\[\frac{ d }{dx }e ^{u}=e ^{u}*\frac{ du }{ dx }\]

Answer 68

You will see this again and again in Calculus problems.

Answer 69

ok if I have e^2x = e^2 ?

Answer 70

therefore, the derivative of e^(2x) is\[\frac{ e }{ dx}e ^{2x}=e ^{2x }*\frac{ d }{ dx}(2x)=?\]

Answer 71

Your tentative answer was partially correct, but you didn't apply the chain rule (to find the derivative of 2x).

Answer 72

can you give me some example ?

Answer 73

Sure. \[\frac{ d }{ dx }e ^{2x}=e ^{2x}*\frac{ d }{ dx }(2x)=e ^{2x}2 = 2e ^{2x}\]

Answer 74

Now, follow that example closely and find the derivative of e^(3x).

Answer 75

the derivative of e^x is simply e^x, or (e^x)*1;
the derivative of e^(2x) is e^(2x)*2;
the derivative of e^(3x) is ????

Answer 76

e^(3x)*3 ?

Answer 77

Very nice. Thanks.
Now, try finding the derivative of e^(sin x), following the same rules. Hint: in the end you must differentiate sin x.

Answer 78

e^(sinx)(cosx) ?

Answer 79

Perfect! You've gotten it!

Answer 80

what is the derivative of \[e ^{\sin ^{2}x}?\]follow the same rules as before. But now you must also apply the power rule, as well as the exponential function, trig function and chain rules.

Answer 81

BRB (be right back)

Answer 82

e^sin2x(2cos^1x) ?

Answer 83

Did you find the derivative of (sin x)^2? Remember the power rule with chain rule.

Answer 84

or
e^sin2x 2sinx ?

Answer 85

Yes, that's a good start; you must now find the derivative of sin x and multiply your previous result by that derivative.

Answer 86

e^sin2x (2sinx)

Answer 87

\[\frac{ d }{ dx }e ^{\sin ^{2}x}=e ^{\sin ^{2}x}(2sinx)\frac{ d }{ dx }\sin x=?\]

Answer 88

So your result is correct, except that y ou have not yet found the derivative of sin x (which is required by the chain rule).

Answer 89

e^sin2x (2sinx)(cosx)

Answer 90

that's great. You're on your way.

Answer 91

thanks, so y=e^sin2(lnx)
e^sin2(lnx) 2sin(lnx) cos(lnx) (1/x) ?

Answer 92

Except that you should label your answer with "(dy/dx)", that's great. Congrats!

Answer 93

dy/dx= e^sin2(lnx) 2sin(lnx) cos(lnx) (1/x) ?

Answer 94

Great.

Whew!

Answer 95

yes yes :) :)

thanks a lot :)

Answer 96

I'd bet you're anxious to move on to other problems or other things.
But @loser66 had a good point: if you were to take the natural log of both sides of the original, given equation, you'd get\[\ln y = (\sin \ln x)^2\]and finding the derivative (dy/dx) of that would be faster and possibly involve fewer steps than those we've gone through.

Answer 97

It's up to you: move on to some other problem, or explore this alternative approach.