Deriving Parameter^x wrt x

**emterics90** · Jun 30th 2007, 08:00 AM

I have $y=7^x$ and I want to find $\frac{dy}{dx}$ . How do I do this? I have tried looking at Google but for some reason I can't find anything. It seems simple but I don't recall ever learning it.

**Jhevon** · Jun 30th 2007, 08:05 AM

Originally Posted by emterics90

I have $y=7^x$ and I want to find $\frac{dy}{dx}$ . How do I do this? I have tried looking at Google but for some reason I can't find anything. It seems simple but I don't recall ever learning it.

A rule of derivatives says:

$\frac {d}{dx} a^x = \ln a \cdot a^x$ , where $a>0$ is a constant

forget google, that should be somewhere in the back of your calculus textbook, they usually have formula sheets there (or some put it in the front, mine has formula sheets in both places)

**Plato** · Jun 30th 2007, 08:07 AM

$b > 0\quad \& \quad y = b^x \quad \Rightarrow \quad y' = b^x \ln (b)$

**emterics90** · Jun 30th 2007, 08:10 AM

Ok, thanks!

**Pterid** · Jun 30th 2007, 03:49 PM

You may be interested (or required by your course!) to know why this rule works. You can derive it using 'implicit differentiation':

Start with the expression

$y = b^x$

and take logarithms of both sides:

$\ln y = \ln (b^x) = x \ln b$

Now, just differentiate both sides of this equation with respect to $x$ :

$\frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln b)$

ie. $\frac{1}{y}\frac{dy}{dx} = \ln b$ (using the chain rule)

$\frac{dy}{dx} = y \ln b = b^x \ln b$

And as a side note: We must have $b>0$ simply because, when working with real numbers, the logarithm of a negative number (or zero) is not defined.

**Krizalid** · Jun 30th 2007, 08:19 PM

$ \begin{aligned} f(x)&=b^x\\ f(x)&=e^{x\ln{b}}\\ f'(x)&=b^x\ln{b}~\blacksquare \end{aligned} $

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