Derivative of a raised to power of x with respect to x


$\dfrac{d}{dx} \, a^x = a^x \log_{\displaystyle e} a$


We know that

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

Use this rule to find the derivative of $a$ raised to the power of $x$ in calculus.

$$\dfrac{d}{dx} \, a^x = \lim_{h \to 0} \dfrac{a^{(x+h)} -a^x}{h}$$

Use product rule of exponents to split the function $a^{(x+h)}$ into two multiplying exponential terms having same base.

$$\implies \dfrac{d}{dx} \, a^x = \lim_{h \to 0} \dfrac{a^x \times a^h -a^x}{h}$$

Take $a^x$ term common from two terms in the numerator.

$$\implies \dfrac{d}{dx} \, a^x = \lim_{h \to 0} \dfrac{a^x (a^h -1)}{h}$$

$$\implies \dfrac{d}{dx} \, a^x = \lim_{h \to 0} a^x \times \dfrac{(a^h -1)}{h}$$

$$\implies \dfrac{d}{dx} \, a^x = a^x \, \lim_{h \to 0} \dfrac{a^h -1}{h}$$

Transformation of the limits function

Take $a^h -1 = t$, then $a^h = 1+t$. Now, transform $h$ terms in terms of $t$ by using its replacing rules. $\dfrac{a^h -1}{h} = \dfrac{t}{h}$. The denominator is still in terms of $h$.

So, use the condition $a^h = 1+t$ and take logarithm both sides. $\log_{\displaystyle e} (a^h) = \log_{\displaystyle e} (1+t)$.

$\implies h \log_{\displaystyle e} a = \log_{\displaystyle e} (1+t)$

$\implies h = \dfrac{\log_{\displaystyle e} (1+t)}{\log_{\displaystyle e} a}$

Now, replace the term $h$ by its value in the function.

$\dfrac{a^h -1}{h} = \dfrac{t}{h} = \dfrac{t}{\dfrac{\log_{\displaystyle e} (1+t)}{\log_{\displaystyle e} a}}$

$\implies \dfrac{a^h -1}{h} = \dfrac{t \times \log_{\displaystyle e} a}{\log_{\displaystyle e} (1+t)}$

$\implies \dfrac{a^h -1}{h} = \dfrac{t \log_{\displaystyle e} a}{\log_{\displaystyle e} (1+t)}$ function.

The function in terms of $h$ is successfully transformed as the function in terms of $t$.

Now it is time to change the limit of the function. We know that $h$ tends to $0$, which means $h \to 0$ then $a^h \to a^0$. Therefore $a^h \to 1$. Subtract $-1$ both sides then $a^h -1 \to 1 -1$ and it is $a^h -1 \to 0$ but $a^h -1 = t$. According to this, $t$ tends to $0$ and it is written as $t \to 0$.

Now express the limits function in terms of $t$ purely.

$$\implies \dfrac{d}{dx} \, a^x = a^x \, \lim_{h \to 0} \dfrac{(a^h -1)}{h} = a^x \lim_{t \to 0} \dfrac{t \log_{\displaystyle e} a}{\log_{\displaystyle e} (1+t)}$$

$$\implies \dfrac{d}{dx} \, a^x = a^x \, \lim_{t \to 0} \log_{\displaystyle e} a \times \dfrac{t}{\log_{\displaystyle e} (1+t)}$$

$$\implies \dfrac{d}{dx} \, a^x = a^x \log_{\displaystyle e} a \, \lim_{t \to 0} \dfrac{t}{\log_{\displaystyle e} (1+t)}$$

Use logarithmic series to expand $\log_{\displaystyle e} (1+t)$

$$\implies \dfrac{d}{dx} \, a^x = a^x \log_{\displaystyle e} a \, \lim_{t \to 0} \dfrac{t}{t -\dfrac{t^2}{2} + \dfrac{t^3}{3} -\dfrac{t^4}{4} \cdots }$$

The literal $t$ is common in each term of the series and take it out common.

$$\implies \dfrac{d}{dx} \, a^x = a^x \log_{\displaystyle e} a \, \lim_{t \to 0} \dfrac{t}{t \Bigg[1 -\dfrac{t}{2} + \dfrac{t^2}{3} -\dfrac{t^3}{4} \cdots \Bigg]}$$

$$\implies \require{cancel} \dfrac{d}{dx} \, a^x = a^x \, \log_{\displaystyle e} a \lim_{t \to 0} \dfrac{\cancel{t}}{\cancel{t} \Bigg[1 -\dfrac{t}{2} + \dfrac{t^2}{3} -\dfrac{t^3}{4} \cdots\Bigg]}$$

$$\implies \dfrac{d}{dx} \, a^x = a^x \log_{\displaystyle e} a \, \lim_{t \to 0} \dfrac{1}{1 -\dfrac{t}{2} + \dfrac{t^2}{3} -\dfrac{t^3}{4} \cdots}$$

Now substitute $t$ is equal to $0$ to release the function from limit.

$\implies \dfrac{d}{dx} \, a^x = a^x \log_{\displaystyle e} a \times \dfrac{1}{1 -\dfrac{0}{2} + \dfrac{0^2}{3} -\dfrac{0^3}{4} \cdots}$

$\implies \dfrac{d}{dx} \, a^x = a^x \log_{\displaystyle e} a \times \dfrac{1}{1 -0 + 0 -0 \cdots}$

$\implies \dfrac{d}{dx} \, a^x = a^x \log_{\displaystyle e} a \times \dfrac{1}{1}$

$\implies \dfrac{d}{dx} \, a^x = a^x \log_{\displaystyle e} a \times 1$

$\therefore \,\,\,\,\, \dfrac{d}{dx} \, a^x = a^x \log_{\displaystyle e} a = a^x \ln a$

It is proved that the derivative of $a$ raised to the power of $x$ with respect to $x$ is equal to $a$ raised to the power of $x$ times $\log_{\displaystyle e}$ of $a$.


$\dfrac{d}{dx} \, 6^x = 6^x \log_{\displaystyle e} 6$

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