- #1

- 79

- 0

u = 1 + tan t

y = u^(1/3)

dy/dt = dy/du x du/dt

u=1+tan t

1/3 u^(-2/3) when u = 1 + tan t x sec^(2)t =

= sec^(2)t/3(1+tan t)^(2/3)

Did I do this correct??

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter helpm3pl3ase
- Start date

- #1

- 79

- 0

u = 1 + tan t

y = u^(1/3)

dy/dt = dy/du x du/dt

u=1+tan t

1/3 u^(-2/3) when u = 1 + tan t x sec^(2)t =

= sec^(2)t/3(1+tan t)^(2/3)

Did I do this correct??

- #2

- 1,236

- 1

[tex] \frac{du}{dt} = sec^{2} t [/tex]

So it should be [tex] \frac{1}{3}(1+ \tan t)^{-\frac{2}{3}}\sec^{2}t [/tex]

- #3

- 131

- 0

If we wish to calculate the derivative of (1) respect to x, q>0 and real

Share:

- Replies
- 2

- Views
- 628

- Replies
- 5

- Views
- 580

- Replies
- 6

- Views
- 845

- Replies
- 2

- Views
- 567

- Replies
- 5

- Views
- 858

- Replies
- 2

- Views
- 780

- Replies
- 2

- Views
- 946

- Replies
- 6

- Views
- 861

- Replies
- 1

- Views
- 1K

- Replies
- 4

- Views
- 854