Question

find dy/dx= 9tan(x/3)

Answer 1

\[\Large y\quad=\quad 9\tan\left(\frac{1}{3}x\right)\]Mmm so we need to find y'?

Answer 2

Do you recall the derivative of tan x?

Answer 3

sec^2 x?

Answer 4

good good c:

Answer 5

Ok we'll get a similar result here, we'll just need to remember to apply the chain rule:\[\Large y'\quad=\quad 9\sec^2\left(\frac{1}{3}x\right)\color{royalblue}{\left(\frac{1}{3}x\right)'}\]Chain rule tells us to multiply by the derivative of the inner function.
So we still need to take the derivative of the blue part.
Understand why we ignore the 9 when we take derivative of tan?

Answer 6

yes (:

Answer 7

is it just 1/3?

Answer 8

The blue part?
1/3x changes to 1/3 when we differentiate?
Yup sounds right! :)

Answer 9

Don't forget to simplify! :O

Answer 10

so then what happens to the front? :o

Answer 11

So you got this far?\[\Large y'\quad=\quad 9\sec^2\left(\frac{1}{3}x\right)\left(\frac{1}{3}\right)\]Remember that multiplication is commutative, we can multiply in any order.
So we can drag this 1/3 to the front and multiply it with the 9 to simplify.

Answer 12

y prime= 3 sec^2 (1/3x) >?

Answer 13

lol minues the >

Answer 14

Yah looks good! without the bird beak.
yay team!\c:/

Answer 15

Yay! :D Mind helping me on a similar one? (:

Answer 16

Sure

Answer 17

would y= sin^2(3x-2) = yprime= 3cos ^2(3x-2) ? ._.

Answer 18

Woops! Let's make sure we understand what the `outermost` function is.\[\Large y\quad=\quad \sin^2(3x-2)\quad=\quad \left[\sin(3x-2)\right]^2\]

Answer 19

With trig functions, we write the exponents in a weird location, so it can be a little difficult to identify the outer function sometimes.

So clearly here, our outer function is ( )^2, right?
So we need to start by applying the power rule.

Answer 20

So power rule, then chain rule:\[\Large y'\quad=\quad 2\left[\sin(3x-2)\right]\color{royalblue}{\left[\sin(3x-2)\right]'}\]

Answer 21

so instead the front would be 6? :)

Answer 22

Yes a 6, but we also get some other business going on :o
Does that last step make sense?
It's a bit tricky.

Answer 23

no :o whats with that? lol

Answer 24

This is an important little tidbit to remember:\[\Large \sin^2x\quad=\quad (\sin x)^2\]^ Do you understand how to take the derivative of this function just above?

Answer 25

yes just bring it down?

Answer 26

2 cos x?

Answer 27

Mmm not quite.
We start by bringing the 2 down, as you said. `But we leave the inside alone while we're changing the outside`.

\[\Large 2(\sin x)\]But chain rule tells us we need to multiply by the derivative of the inside:\[\Large 2(\sin x)\color{royalblue}{(\sin x)'}\]
Are you starting to see the difference?
We can't apply the power rule and derivative of sine at the same time.
Outer function is ( )^2, we apply power rule.
Then we multiply (on the outside) by the derivative of the inner part.

Answer 28

Uh oh :( I heard that...
the sound of your head esploding... ugh

Answer 29

lol (: no i guess i kind of get it

Answer 30

kinda

Answer 31

Umm here's another example that might help.\[\Large (3x+1)^2\]If we take the derivative, it doesn't change to:\[\Large 2(3)\]When we bring the 2 down, we should still have (3x+1) inside, right?
We can't apply the power rule AND the 3x+1 at the same time. It leads us down the wrong path.

Answer 32

Here's the derivative of this example just to compare:\[\Large \left[(3x+1)^2\right]'\quad=\quad 2(3x+1)(3)\]

Answer 33

oh okay(: i get that one!

Answer 34

so the equation would be set up as 2 sin(3x-2) x 2cos(3x-2) x 3 ?

Answer 35

ok my head is exploding lol

Answer 36

Ahhh you're so close, you have an extra 2 for some reason.\[\Large 2\sin(3x-2)\]Then we multiply by the derivative of the inner function (which in this case is sin(3x-2) which becomes cos(3x-2) ):\[\Large 2\sin(3x-2)\cos(3x-2)\]
Then we multiply by the derivative of the inner function again
(which in this case is (3x-2) which becomes (3) ):
\[\Large 2\sin(3x-2)\cos(3x-2)(3)\]

Answer 37

ohhhh okay

Answer 38

got that

Answer 39

6 sin(3x-2)cos(3x-2) ?

Answer 40

Yes

Answer 41

thank you :)