help with derivatives

[integragirl]

I have two derivatives that I need help with please. The first one is y=(4x^3)(4^3x) the x is part of the exponent. The second one states y=ln(arctan 5x). Any help would be greatly appreciated!

 

[stapel]

To clarify, are your questions as follows...?

. . . . .\(\displaystyle \large{\mbox{1) Differentiate\,}\,y\,= \,4x^3 4^{3x}}\)

. . . . .\(\displaystyle \large{\mbox{2) Differentiate\,}\,y\,= \ln{(\arctan{(5x)}})}\)

If so, what have you tried? How far have you gotten? Where are you stuck?

If not, please reply with clarification.

Thank you.

Eliz.

 

[skeeter]

for y=(4x^3)(4^3x) ...

use the product rule, and a hint ... d/dx(a^u) = (1/lna)(a^u)(du/dx)

for y=ln(arctan 5x) ...

if y = ln[f(u))], the y' = [f'(u))/f(u))](du/dx)

 

[integragirl]

To stapel, yes you are correct. For (4x^3)(4^3x) I'm not sure how to deal with the exponent of the second part, and for the other one ln(arctan 5x) I'm not sure if I have to apply the chain rule.

 

[Concorde]

Integ,

a^b is equivilant to e^(b*ln(a))

You should be able to get it from there on out I think.

Oh yeah for the second one yes, you do have to apply the chain rule.

 

[skeeter]

I'm not sure how to deal with the exponent of the second part

did you even see this ?

use the product rule, and a hint ... d/dx(a^u) = (1/lna)(a^u)(du/dx)

and for the other one ln(arctan 5x) I'm not sure if I have to apply the chain rule.

what does this tell you?

if y = ln[f(u)], then y' = [f'(u)/f(u)](du/dx)