A discovery in my book want to prove this: f(x) = b2 then f'(x) = f'(0) * bx
The derivative of y = bx
Let's say b = 2. y = 2x
Complete the table:
| x | y | dy/dx | dy/dx / y |
|---|---|---|---|
| 0 | 1 | 0 | 0 |
| 0.5 | Root(2) | 1 / (2Root(2)) | 1/ 4 |
| 1 | 2 | 1 | 1 / 2 |
| 1.5 | 2Root(2) | 3 / Root(2) | 3 / 4 |
| 2 | 4 | 4 | 1 |
Complete the table for any number b:
| X | y | dy/dx | dy/dx / y |
|---|---|---|---|
| 0 | 1 | 0 | 0 |
| 0.5 | Root(b) | 1 / (2Root(b)) | 1/2b |
| 1 | b | 1 | 1/b |
| 1.5 | b Root(b) | 3 / Root(b) | 3/b2 |
| 2 | b2 | 2b | 2/b |
I still don't see how f(x) = b2 then f'(x) = f'(0) * bx is true...
Isn't the derivative of b0 = 0? So f'(0) = 0 So, f'(x) = f'(0) * bx = 0...
I'm definetely missing something obvious here. What is it? Please help!
Thanks.
+71 Derivative of y = b^x . b is a constant and x is the variable,so differentiate with respect to x,not b.
The result is dy/dx = b^x ln(b). Here is a proof of the result.
Take base e logs of both sides to get
ln(y) = x ln(b)
differentiate both sides with respect to x
(1/y)dy/dx = ln(b) { using chain rule on left hand side }
dy/dx = y ln(b) = b^x ln(b).
In general ,the derivative of a^x = a^x ln(a).
