Question

Derivative question

Answer 1

y = (log base 5(tan3x))(5^(x^3-7))

Hopefully that is clear... the 5 is being raised to (x^3 - 7)

Answer 2

I need to find the derivative

Answer 3

Here is what I have:

(log base 4(tan3x))(3ln(5)x^2)(5^(x^3 - 7)) + (5^(x^3-7))(ln(tan3x)/ln4)

Answer 4

I know, it's a lot... is the derivative of log base 4 of tan3x equal to ln(tan3x)/ln(4) ?

Answer 5

i'll try and latex the question first

Answer 6

latex the question? what does that mean :P

Answer 7

\(y = (\log_5 (tan3x))(5^{x^3-7})\) ??

Answer 8

latex = typeset

Answer 9

but yes, to answer yr question...

\(\dfrac{d}{dx} \left( \log_4 (\tan 3x)\right) = \dfrac{d}{dx} \left( \dfrac{\ln (\tan 3x)}{\ln 4}\right) \)

Answer 10

All right except the base should be 4, sorry

Answer 11

Does my answer to the orig. question look right?

Answer 12

let me do it.

few mins

Answer 13

thanks :)

Answer 14

you're using the product rule but i don't think you have differentiated one of the terms

\(\dfrac{d}{dx} (5^{x^3 - 7}) = (5^{x^3 - 7}) (3x^2) (\ln 5) \) so that bit is right

but \(\dfrac{d}{dx} (\log_4 ( \tan 3x)) = \dfrac{d}{dx} \dfrac{\ln (\tan 3x )}{\ln 4}\)

\(= \dfrac{1}{\tan (3x)} * \dfrac{\sec^2 (3x) * 3}{\ln 4}\)

Answer 15

ie i can't see that last bit in your answer

Answer 16

if zep is here to the rescue, i may bail :-)

Answer 17

haha xD lemme look over my work... one sec

Answer 18

zep to the rescue!

Answer 19

no rush

Answer 20

lol seems like no rescue needed here XD

Answer 21

So \[\ln(\tan3x)/\ln4\] is not the derivative of log base 4 of tan3x? I have to take the derivative of that?

Answer 22

Can you walk me through taking the derivative of that?

Answer 23

i think you did this bit

\(\dfrac{d}{dx} (\log_4 ( \tan 3x)) = \dfrac{d}{dx} \dfrac{\ln (\tan 3x )}{\ln 4}\)

right?

Answer 24

but you don't seem to have done the calculus. so it would appear to me

Answer 25

Yes, but I don't remember how to take the derivative of that... I see you did it above, but how did you get that answer?

Answer 26

ok

step by step

Answer 27

\(\dfrac{d}{dx} \dfrac{\ln (\tan 3x )}{\ln 4}\)

\(\dfrac{1}{\ln 4} \dfrac{d}{dx} \ln (\tan 3x )\)

and now chain rule

\(\dfrac{d}{dx} ( \ln ( f(x) )) = \dfrac{1}{f(x)} f'(x)\)

Answer 28

Wouldn't I need to use the quotient rule?

Answer 29

Oh nevermind I see

Answer 30

so you ens up with

\(\dfrac{1}{\ln 4} * \dfrac{1}{\tan 3x} * \dfrac{d}{dx} ( \tan 3x)\)

Answer 31

and then i'd break up what ever you get once you finish that into sines and cos's and try simplify it a bit

Answer 32

My calc teacher doesn't care about simplifying, she said we can just leave stuff how it is...
so would it be sec^2(3x)*3 over tan(3x) ?

Answer 33

I appreciate the help :D

Answer 34

\[\large\rm \frac{d}{dx}\log_4(\tan3x)\quad=\quad \frac{3\sec^2(3x)}{(\tan3x)\ln4}\]Yes.

Answer 35

Thanks :)

Answer 36

You figure out the other derivative ok? :o

Answer 37

Other derivative? Do you mean the 5^(x^3-7)? already done zeppy

Answer 38

Oh I missed that :P woops