Derivative ^x, is this correct?

  • #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]
 

Answers and Replies

  • #2
867
0
You need to use the product rule. If you didn't know:
[tex]19^x = e^{\ln 19^x} = e^{x\ln 19} = e^{(\ln 19)x}[/tex]

I'm double-checking everything now. :redface:
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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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]
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.

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)) `
 

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