Question

i need help applying chain rule

http://webwork.math.ttu.edu/wwtmp/equations/10/ab7506a44aafa9272098cb86d31c771.png

Answer 1

chain rule is only going to be used for the tan(3x). First differentiate the outside, then the inside, so: Sec^2(3x) * 3 = 3Sec^2(3x)

Answer 2

\[(trig(f(x))'=f'(x)(trig)'(f(x))\]

Answer 3

im not really sure how to apply what you just wrote freckles...

Answer 4

\[(\tan(3x))'=(3x)'(\tan)'(3x)=(3) \sec^2(3x)\]

Answer 5

where does that second 3x come from?

Answer 6

It's like within the tangent:
(3x)' * (Tan(3x))'

Answer 7

wait i actually understand that but what about the rest of the function? do i just plug (3)sec^2(3x) into my quotient rule

Answer 8

yea that's what you plug in into your quotient rule

Answer 9

\[\frac{3\sec^2(3x) \cdot (bottom)-\tan(3x) \cdot (bottom)'}{(bottom)^2}\]

Answer 10

well that gives me ((x^2)(3sec^2x)- (tan(3x))(2x))/((x^2)^2)
and I've tried that answer several times and its not working

Answer 11

\[\frac{3 \sec^2(3x) \cdot x^2-\tan(3x) \cdot 2x}{x^4} =\frac{3x \sec^2(3x)-2 \tan(3x)}{x^3}\]

Answer 12

how did you get to x^3 on bottom and where did the x on the two go? did we just take 1 power of x of the top and bottom?

Answer 13

you know how to factor right?

Answer 14

Each term on top has a factor of x in common

Answer 15

So you can factor x out on top

Answer 16

And bottom top and bottom have a common factor x

Answer 17

we assumed x/x=1

Answer 18

ok i think that makes sense

Answer 19

\[\frac{x ( 3 \sec^2(3x) \cdot x- \tan(3x) \cdot 2)}{x \cdot x^3}\]

Answer 20

See both top and bottom have a common factor x

Answer 21

\[\frac{\not{x} ( 3 \sec^2(3x) \cdot x- \tan(3x) \cdot 2)}{\not{x} \cdot x^3}\]

Answer 22

ok freckles that does make sense