- #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??

- 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

- 1,235

- 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