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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.

 Jul 30, 2017
 #1
avatar+71 
+1

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).

 Jul 30, 2017
 #2
avatar+99388 
+1

If the chain rule confuses you here is another way to look at it.

 

\(y=b^x\\ lny=lnb^x\\ lny=x*lnb\\ x=\frac{lny}{lnb}\\ \frac{dx}{dy}=\frac{1}{ylnb}\\ \frac{dy}{dx}=ylnb\\ \frac{dy}{dx}=b^xlnb\\ \)

.
 Jul 31, 2017

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