Question

dy/dt y=cos [sqrt(4t+11)]

Answer 1

Chain rule.

Answer 2

Let u = √4t+11) --> u' = ....

Answer 3

I got the answer as \[-sinx \sqrt{4t+11} + \cos (2) (4t+11)^{-1/2}\]

Answer 4

but the answer is actually with no (cos), what did I do wrong?

Answer 5

Shouldn't be adding anything.
Chain rule says to multiply all the derivatives together.

Answer 6

( cosu)' = - u' sinu

Answer 7

can you give me all steps so I can compare what I did wrong?

Answer 8

There are three functions:
cos [sqrt(4t+11)],
sqrt(4t+11),
and 4t+11.

Find the derivatives of those three, then multiply them together.

Answer 9

If it helps, think of it like this:
y=cos W
W=sqrt(U)
U=4t+11

Answer 10

it worked thanks

Answer 11

so let's say if it's \[t ^{6}(t ^{5}+4)^{5}\]

How would you divide out each individual like you just did

Answer 12

Option one is to simplify by multiplying everything out then going term-by-term.
Option two is to see that you have a product of U*V :
U = t^6
V= W^5
W = t^5+4

You'll need a combination of product rule and chain rule here.

Answer 13

thanks

Answer 14

how did they get the answer as \[t ^{5}(t ^{5}+4)^{4}(31^{5}+24)\]

Answer 15

For
\[\large \frac{d}{dx} t^6(t^5+4)^5 \space ?\]

Answer 16

\[6t ^{5}(....)\]

Answer 17

I don't understand this either,
\[\large "t^5(t^5+4)^4(31^5+24)"\]
What is that the answer to?

Answer 18

to that same problem, but that is the answer in the solution manual

Answer 19

Then I still don't get it.

Answer 20

u = t^6 ---> u' = 6t^5
v = ( t^5 + 4) ^5 ---->v' = 5( t^5 + 4) ^4 * 5t^4
= 25 t^4 ( t^5 + 4) ^4

Answer 21

,,,,
= t^5 ( t^5 + 4) ^4 [ 6( t^5 + 4) + 25 t^5 ]

Answer 22

= t^5 ( t^5 + 4) ^4 [ 31 t^5 + 24 ]

Answer 23

So the textbook has typo at 31^5 !

Answer 24

@thuyvy Questions?

Answer 25

I got it, thanks :)

Answer 26

you're the best chlorophyll, you saved me !! Thanks :)

Answer 27

@Chlorophyll: can you help me on some other problems?

Answer 28

Did you open new post, keep it simple one question per post will be less confused for the solver!