- #1

- 78

- 1

## Homework Statement

[tex]y=\sqrt{x} * 19^x[/tex]

## Homework Equations

## The Attempt at a Solution

[tex]y=x^{1/2}*19^x[/tex]

[tex]y'=1/2x^{-1/2}*19[/tex]

- Thread starter neutron star
- Start date

- #1

- 78

- 1

[tex]y=\sqrt{x} * 19^x[/tex]

[tex]y=x^{1/2}*19^x[/tex]

[tex]y'=1/2x^{-1/2}*19[/tex]

- #2

- 867

- 0

[tex]19^x = e^{\ln 19^x} = e^{x\ln 19} = e^{(\ln 19)x}[/tex]

I'm double-checking everything now.

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

No, that's not right and I don't see how you could have got that! The derivative of a product is typically a sum of derivatives and the dervative of "[itex]19^x[/itex]" is definitely not 19.## Homework Statement

[tex]y=\sqrt{x} * 19^x[/tex]

## Homework Equations

## The Attempt at a Solution

[tex]y=x^{1/2}*19^x[/tex]

[tex]y'=1/2x^{-1/2}*19[/tex]

I would just take the logarithm of both sides immediately:

[tex]ln(y)= ln(\sqrt{x}19^x)= ln(x^{1/2})+ ln(19^x)[/tex]

[tex]ln(y)= (1/2)ln(x)+ x ln(19)[/tex]

Now differentiate both sides and solve for y'.

- #4

- 327

- 1

y = sqrt(x) * 10^x

y` = sqrt(x) * (10^x)` + 10^x * (sqrt(x)) `

y` = sqrt(x) * (10^x)` + 10^x * (sqrt(x)) `

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 1K

- Last Post

- Replies
- 0

- Views
- 2K

- Last Post

- Replies
- 7

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 885

- Replies
- 3

- Views
- 1K

- Replies
- 2

- Views
- 3K

- Replies
- 4

- Views
- 1K

- Replies
- 15

- Views
- 985

- Replies
- 4

- Views
- 1K