Question

Assume that a>0. Differentiate the function at a^t

Answer 1

ta^(t-1)???

Answer 2

well I thought that too , but no because a is always used to identify an actual number and not a variable

Answer 3

oh whoops didnt notice that give me a sec so the variable is the t???

Answer 4

so assuming that a is a value and t is the variable:
ln(a)a^t

Answer 5

this is one of the rules

Answer 6

thats what I was thinking also
let me check thanks

Answer 7

hey r u there?

Answer 8

sorry i just lost connection for a sec

Answer 9

so was it correct?

Answer 10

Yes, sorry lost connection as well. I thought it was something to that effect, thanks!

Answer 11

hey u r still here. thought u were long gone :D

Answer 12

ok bye for now gonna go study calc

Answer 13

Yes, and you can prove it this way:
\[ a^x = (e^{\ln a})^x \]
because a = e^(ln a)

Now,
\[ (e^{\ln a})^x = e^{(\ln a)x} \]
therefore
\[ \frac{d \ }{dx} a^x = \frac{d \ }{dx} e^{(\ln a)x} = \ln a .e^{(\ln a)x} \]
because ln a is just a constant. Finally,
\[ \ln a .e^{(\ln a)x} = \ln a . a^x \]
therefore
\[ \frac{d \ }{dx} a^x = \ln a . a^x \]

Answer 14

james always gives the best answer. TRUST JAMES :D