Question

use logathrimic differentiation to differentiate the following:
y = 3 ^ tanx

thanks for any help

Answer 1

log and diff ... right?

Answer 2

Take the natural log of both sides and use properties of logs to get
\[\ln(y)=\tan(x)\ln(3)\]
ln(3) is a constant, so when you differentiate, you can treat it like any other constant.
\[\ln(y)'=(\ln(3)\tan(x))'\]
Pull out the ln(3)\[\frac{1}{y}=\ln(3)(\tan(x))'\]\[\frac{1}{y}=\ln(3)\sec^2(x)\]
Flip it and you're done.

Answer 3

ln (y = 3 ^ tanx)

lny = ln 3 ^ tanx

lny = tanx ln 3

y'/y = tan'x ln 3 + tanx ln' 3

y'/y = sec^2 x ln 3

y' = y(sec^2 x ln 3)

Answer 4

ln3 is a constant, so you don't need to use chain rule, I think. It should go to 0 anyway.

Answer 5

since constants go to zero; the chain rule works on them and they zero out; and your left with the so called "constant" rule

Answer 6

i didnt know whether ln 3 went to 1/3..like most ln derivatives do..thanks so much for the quick reply

Answer 7

[2x]' = 2x'+2'x

[2x]' = 2+0x

[2x]' = 2

Answer 8

Right, what I meant by the going to 0 part. Where is the y coming from?

Answer 9

ln3 is just a constant; no need to try to make it a variable ;)

Answer 10

implicit differentation; y -> y'

Answer 11

Oh, right, duh. Thanks! Blonde moment.

Answer 12

ln(y) chains to; 1/y * y'

Answer 13

thanks so much for the help

Answer 14

yw