Question

what is the derivative of 1/2 sec^2 3x(tan^2 3x - 1)

Answer 1

\[\Large \frac{1}{2}\sec^2(3x)\tan^2(3x-1)\]Is this what the function looks like?

Answer 2

yes please HELP

Answer 3

So we have the `product` of two functions of x, what rule for differentiation should we use? :)

Answer 4

i do't know :o
i'm up from 7 : 30 in the evening til 5:56 in the morning i'm doing my take home exam + stat problem + review for chemistry
my head is not functioning :(

Answer 5

We have a product, we should apply product rule silly! :)

Answer 6

Here is our setup for product rule:\[\large =\color{royalblue}{\left[\frac{1}{2}\sec^2(3x)\right]'}\tan^2(3x-1)+\frac{1}{2}\sec^2(3x)\color{royalblue}{\left[\tan^2(3x-1)\right]'}\]

Answer 7

ohh that's it HAHAHa sorry

Answer 8

We'll need to take the derivative of the blue parts.

Answer 9

that's vdu + udv ,am i right ?

Answer 10

yes :)

Answer 11

Make sure you're comfortable with the exponent notation on trig functions:\[\Large \sin^2x\quad=\quad (\sin x)^2\]

Answer 12

hmm

Answer 13

2 sec 3x ?

Answer 14

6 tan (3x - 1)?

Answer 15

No no, slow down :o

Answer 16

So let's look at the first blue term a sec:\[\Large \color{royalblue}{\left[\frac{1}{2}\sec^2(3x)\right]'}\]
You have the right idea, we start by applying the power rule,\[\Large \color{royalblue}{\left[\frac{1}{2}\sec^2(3x)\right]'}\quad=\quad \frac{1}{2}\cdot 2 \sec(3x)\color{royalblue}{\left[\sec(3x)\right]'}\]But the chain rule tells us we have to multiply by the derivative of the inner function ( that's where the new blue part is coming from, chain rule).

Answer 17

hmm.

Answer 18

The outermost function was ( stuff )^2

We applied the power rule which gave us 2( stuff ),
but the chain rule tells us we also have to multiply by the derivative of the stuff.

=2( stuff )( stuff )'

Answer 19

Chain rule is always a tricky one to get down :(

Answer 20

can you please give me the answer and i'll just figure out how you got them ?

Answer 21

you -_-

Answer 22

I'll do the first blue term for you, maybe it will help you see how to do the other one :3

Answer 23

i'll be having my statistics later and my finals exam is 6 hours after

Answer 24

\[\Large \color{royalblue}{\left[\frac{1}{2}\sec^2(3x)\right]'}\]Applying power rule then chain rule gives us,\[\Large =\quad \frac{1}{2}\cdot2\sec(3x)\color{royalblue}{\left[\sec(3x)\right]'}\]Taking the derivative of secant gives us secant tangent,\[\Large =\quad \frac{1}{2}\cdot2\sec(3x)\left[\sec(3x)\tan(3x)\right]\color{royalblue}{(3x)'}\]
We have to apply the chain rule again, multiplying by the derivative of the innermost function (3x), giving us,\[\Large =\quad \frac{1}{2}\cdot2\sec(3x)\left[\sec(3x)\tan(3x)\right](3)\]

Answer 25

We can simplify this down by moving the coefficients to the front,\[\Large =\quad 3 \sec(3x)\sec(3x)\tan(3x)\]And further by multiplying the secants together,\[\Large 3\sec^2(3x)\tan(3x)\]

Answer 26

So there's the derivative of our first blue portion, yay!

Answer 27

then the second one ? :)

Answer 28

-_-

Answer 29

Since we found the derivative of the outside function, sec(3x) in this case, we need to find the derivative of the inside function, which is 3x

the derivative of 3x is 3. and we get

2(sec(3x))sec(3x)tan(3x)3

We simplyfy the whole equation, multiply 2 and 3....and multply the 2 sec(3x)'s and we get

6(sec(3x))^2tan(3x)...we can further simplyfy to make it neater and we get

6sec^2(3x)tan(3x)
this is the first part right ?

Answer 30

The outside function was the square,
the function inside of that was the sec(3x),
the function inside of that was (3x).

So it required us to do chain rule several times. D:

I think maybe you're forgetting the 1/2 in front or no? :o
Looks ok besides that though.