The derivative of 4^x + 3^x + 9^-x would be zero.

[EvilPony]

if anyone can teach me how to do this that would be great, thanks.

dereiv. means derivative sorry

 

[stunner5000pt]

if anyone can teach me how to do this that would be great, thanks.

dereiv. means derivative sorry

when I am doing something with the form something raised to x i always remember - keep the tern , log (or ln, same meaning here) the base number

and then differentiate the exponent

for example for \frac{d}{dx} (3^x) = 3^x Log3 (1)

as you can see keep the function 3^x, log the base Log3, and then differentiate teh numerator (1).

 

[Zurtex]

All I can say to the above post, is eh? That would mean that it would be 0. In general:

\frac{d}{dx} \left( a^x \right) = \ln (a) \; a^x

Where a is some constant. Here is the method used to work it out and generally useful for this type of problem:

y= a^x

\ln y = \ln \left( a^x \right)

\ln y = x \ln a

\frac{dy}{dx} \frac{1}{y} = \ln a

\frac{dy}{dx} = (\ln a)y

\frac{d}{dx} \left( a^x \right) = \ln (a) \; a^x

 

[stunner5000pt]

All I can say to the above post, is eh? That would mean that it would be 0. In general:

\frac{d}{dx} \left( a^x \right) = \ln (a) \; a^x

Where a is some constant. Here is the method used to work it out and generally useful for this type of problem:

y= a^x

\ln y = \ln \left( a^x \right)

\ln y = x \ln a

\frac{dy}{dx} \frac{1}{y} = \ln a

\frac{dy}{dx} = (\ln a)y

\frac{d}{dx} \left( a^x \right) = \ln (a) \; a^x

what would be zero??