Question

`lim_(h->0)(25 xx 5^h-25)/h =` ?

Answer 1

`f(x) = 5^x`
`f(a) = 25`
`color(white)"----"a = 2`

Explanation:

This is a limit definition of a derivative.

The formula for this is:

`lim_(h->0)(f(a+h) - f(a))/h`

This tells us that

`f(a+h) = 5^(2+h)`

`f(a) = 25`

How can we figure out what `f(x)` is? Well, in `f(a+h)`, we can see that the `h` is in the exponent, so there's a good chance that the function is some sort of exponential function, probably with a base of 5 as well. So, let's try this:

Assume `f(x) = 5^x`

This means that

`f(a+h) = 5^(2+h)`

`a+h = 2+h`

`a = 2`

Now to check if this works out correctly.

`f(a) = f(2) = 5^2 = 25`, which we know is true from earlier.

Therefore, we can say that `f(x) = 5^x` and `f(a) = 25`.

Final Answer

Answer 2

See below.

Explanation:

Calling

`f(x) = 5^x` we have

`lim_(h->0)(f(2+h)-f(2))/h = f'(2) = log_e 5*5^2= 25*log_e 5`

NOTE:

As we know `d/(dx)e^(alpha x) = alpha e^(alpha x)` and

`a= e^(log_e a)` then `a^x = e^(log_e a * x)` and then

`d/(dx) a^x = d/(dx)e^(log_e a * x) = log_e a *e^(log_e a * x) = log_e a*a^x`