Proving the Derivative of $f'(x) = b^x$

Question

Given $f(x) = b^x = e^{x\ln b}$ for $b > 0$, can someone show me how $f'(x) = \ln b e^{x\ln b}$ ?

Answer 1

As you said: $$f(x)=b^x\\\ln f(x)=x\ln b \\f(x)=e^{x\ln b}$$

Now:

$$\frac{d}{dx}e^u=e^u \frac{du}{dx}$$

Hence:

$$u=x\ln b \\ \frac{du}{dx}=\ln b$$

So:

$$f'(x)=e^u \frac{du}{dx}=e^{x\ln b}\ln b$$

Answer 2

$(e^u)'=u'e^u$. Then, if you set $u(x)= x\ln b$, we have $u'(x)=\ln b$. If you apply the formula, you arrive at $$\ln be^{x\ln b}$$

Answer 3

$$f'(x) = b^xln(b)$$ $$f'(x) = e^{ln(b)x}ln(b)$$

Answer 4

The definition for the derivative of a real valued function $f$ is $$\frac{df}{dx}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$

Letting $f(x) = e^{x \ln b }$, we have \begin{eqnarray} \frac{df}{dx}&=&\lim_{h\to 0}\frac{e^{(x+h) \ln b }-e^{x \ln b }}{h}\\ &=& \lim_{h\to 0}\frac{e^{x \ln b }(e^{h\ln b}-1)}{h}\\ & = & e^{x \ln b }\lim_{h\to 0}\frac{b^{h}-1}{h} \\ & = &e^{x \ln b }\ln b. \end{eqnarray}