A discovery in my book want to prove this: **f(x) = b ^{2} **then

The derivative of y = b^{x}

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/b^{2} |

2 | b^{2} | 2b | 2/b |

I still don't see how **f(x) = b ^{2} **then

Isn't the derivative of b^{0} = 0? So **f'(0) = 0 **So, **f'(x) = f'(0) * b ^{x} **= 0...

I'm definetely missing something obvious here. What is it? Please help!

Thanks.

Guest Jul 30, 2017

#1**+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).

frasinscotland
Jul 30, 2017