Posted by **Raj** on Wednesday, June 13, 2007 at 10:40pm.

What is wrong with the following reasoning???

f(x) = 2^x

f'(x) = x[2^(x-1)]

Although n x^(n-1) is the derivative of x^n, you cannot apply similar rules when x is the exponent and the number being raised to a power is a constant.

The derivative of b^x, where b is a constant, is ln(b) * b^x

For a proof, see

http://math2.org/math/derivatives/more/b^x.htm
## Answer This Question

## Related Questions

- Math-Calculus - Hi, I am trying to figure out what the limit as h approaches 0 ...
- Math 9 - use exponent rules to simplify. Write as a single power, don't need to...
- Calculus Derivative - I'm trying to calculate the derivative of this: P= mv/ ...
- Calculus - If I have a function f(x), and am given its derivative, f'(x): may I ...
- CALCULUS - what is the derivative of ln(x-1)? If you are studying the calculus ...
- Calculus - f(x)=sin^(7)x The 7 is an exponent and the x is not. Find the ...
- calculus - how do you find the derivative of (X^2(2+x^1/2))/x^4. I'm confused ...
- applied calculus - 1. Consider the function y = xx (for x> 0). a) Why does ...
- Algebra - How do you simplify the following and please show all work. (-3x(-4 ...
- Pre calculus - How do I solve this exponent problem? I understand the exponent ...

More Related Questions