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Given $f(x) = 5^{3x}$. Find $f'(x)$ using definition of a derivative.

The definition of the derivative of $f(x)$ is

$f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$

The derivative of $f(x) = 5^{3x}$ is

$f'(x) = \lim_{h \to 0} \dfrac{5^{3(x + h)} - 5^{3x}}{h}=\lim_{x \to 0} \dfrac{5^{3x}5^{3h} - 5^{3x}}{h}=\lim_{x \to 0} \dfrac{5^{3x}\left(5^{3h} - 1\right)}{h}$

And this is where I get stuck. Is there any way to proceed from here appropriately?

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  • 2
    $\begingroup$ It should say $\lim_{h\to 0}$ not $x$ $\endgroup$ – Rustyn May 5 '14 at 4:04
  • $\begingroup$ This should be as $h$ approaches zero. $\endgroup$ – EgoKilla May 5 '14 at 4:04
  • $\begingroup$ Correctedㅤㅤㅤㅤㅤㅤ $\endgroup$ – Hero May 5 '14 at 4:24
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What you need to show is that: $$ \lim_{h\to 0} \frac{5^{3h}-1}{h}=\lim_{h\to 0} \frac{125^{h}-1}{h}=\ln(125) $$ Note, we have the following: $$\lim_{x\to 0} \frac{\log_a(1+x)}{x} = \lim_{x\to 0} \frac{\log_e(1+x)}{x\log_e{a}}=\lim_{x\to0} \frac{\sum_{i=1}^{\infty} (-1)^{n+1} \frac{x^{n}}{n}}{x\log{a}} =\lim_{x\to 0} \frac{1+\sum_{i=2}^{\infty} (-1)^{n+1} \frac{x^{n}}{n}}{\log{a}} = \frac{1}{\log{a}} = \frac{1}{\frac{\log_a{a}}{\log_{a}{e}}} = \log_{a}{e}$$

Now set $y=a^x -1.$ Hence $a^x = 1+y$ and $$ x = \log_a(1+y) $$ As $x\to 0; y\to 0$
Therefore: $$ \lim_{x\to 0}\frac{a^x-1}{x}=\lim_{y\to 0}\frac{y}{\log_a(1+y)}=\log_e a $$ This is taken from my answer here: Proving that $\lim_{h\to 0 } \frac{b^{h}-1}{h} = \ln{b}$

You might look at some other approaches there.

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