Question

help?

Answer 1

\[(\sec(x))^{5x}\]
find dy/dx

Answer 2

What's up with you bozo's that type HELP in the description :P I hate that! lol

Answer 3

Hmm we have x in the BASE, and x in the exponent :O
Uh oh, can't apply power rule to this one.

Need to use logarithmic differentiation.

Answer 4

\[\huge y=(\sec x)^{5x}\]Taking the natural log of both sides gives us,\[\huge \ln y=\ln\left[(\sec x)^{5x}\right]\]

Answer 5

Using a familiar rule of exponents,\[\huge \log(b^a)=a\cdot \log(b)\]will allow us to get the 5x OUT of the exponent location.

Answer 6

cause i'm such a bozo, you haven't noticed? ;)

Answer 7

\[\huge \ln y=5x\cdot \ln\left(\sec x\right)\]

Answer 8

From here you'll take the derivative of both sides, getting some fun stuff on both sides.

Answer 9

hahaha! let me try.. I do the chain rule for ln(sec(x)) right? then product rule with that derivative and 5x?

Answer 10

Hmm you might have that process backwards, always product rule setup first. Unless I'm simply misunderstanding you :D

Answer 11

I didn't know that... rreallyyy?

Answer 12

Ummmmmmmmmmmmmmmm

Answer 13

well actually, don't you sorta do them at the same time..?

Answer 14

I Dunno :P You just apply which ever rule is staring you in the face at that moment. Right now, we have a product of 2 functions.
Taking the derivative of the right side will give us this setup,\[\huge 5x\cdot \left[\ln\left(\sec x\right)\right]'+(5x)'\cdot \ln(\sec x)\]

Answer 15

Yah I suppose you do do do do do do do them at the same time. :)
I just generally think of the product rule SETUP as coming first.

Answer 16

because you have to find the derivative of g(x) to complete the product rule and to find g(x) you do the chain rule

Answer 17

Although you probably do the setup in your head, not on paper :O So yah, same time-ish

Answer 18

nope, paper always :) just in case you make MISTAKES like my post before ;p

Answer 19

ah :)

Answer 20

I got.. (5)(ln(sec(x)))+(5x)(tan(x))

Answer 21

Yah that sounds correct for the right side.

Answer 22

What do you get on the left side? with the y and the log and the thing and the thing?

Answer 23

ln(y) the derivative is 1/y

Answer 24

We're taking the derivative with respect to X, so a y' should pop out also.

Answer 25

1/x?

Answer 26

wait...

Answer 27

\[\huge (\ln y)'=\frac{1}{y}y'\]

Answer 28

don't I replace what y is?

Answer 29

After you solve for y', yes.

Answer 30

? ^ im confused with that

Answer 31

You have y'(1/y) on the left. To solve for y', multiply both sides by y.

Answer 32

oh!!!

Answer 33

I got it thanks!

Answer 34

yay