Proofs for the derivatives of eˣ and ln(x)

The derivatives of eˣ and ln(x) are two important results in calculus. Their proofs explain how these derivative formulas are obtained using the basic rules of differentiation.

Proof for the Derivative of e^x

Using the first principle of derivatives,

f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

Let, f(x)=e^x.

Then, \begin{aligned} f'(x) &=\lim_{h\to0}\frac{e^{x+h}-e^x}{h}\\[6pt] &=\lim_{h\to0}\frac{e^x\cdot e^h-e^x}{h}\\[6pt] &=\lim_{h\to0}\frac{e^x(e^h-1)}{h}\\[6pt] &=e^x\lim_{h\to0}\frac{e^h-1}{h} \end{aligned}

From the fundamental limit, \lim_{h\to0}\frac{e^h-1}{h}=1.

Therefore, \boxed{f'(x)=e^x.}

Example 1: Find the derivative of y = {e^{ – {x^3}}}?

Solution:

By the chain rule,

{y^\prime = \left( {{e^{ – {x^3}}}} \right)^\prime }={ {e^{ – {x^3}}} \cdot \left( { – {x^3}} \right)^\prime }={ {e^{ – {x^3}}} \cdot \left( { – 3{x^2}} \right) }={ – 3{x^2}{e^{ – {x^3}}}.}

Example 2: Find the derivative of y =\frac{{{e^x} – 1}}{{{e^x} + 1}}?

Solution:

Use here the quotient rule:

{y^\prime = \left( {\frac{{{e^x} – 1}}{{{e^x} + 1}}} \right)^\prime }={ \frac{{{e^x}\left( {{e^x} + 1} \right) – \left( {{e^x} – 1} \right){e^x}}}{{{{\left( {{e^x} + 1} \right)}^2}}} }={ \frac{{{\cancel{e^{2x}}} + {e^x} – \cancel{e^{2x}} + {e^x}}}{{{{\left( {{e^x} + 1} \right)}^2}}} }={ \frac{{2{e^x}}}{{{{\left( {{e^x} + 1} \right)}^2}}}.}

Proof for the Derivative of ln(x)

Using the first principle of derivative, f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

Let, f(x)=\ln(x)

Substitute f(x)=\ln(x) into the formula:

f'(x)=\lim_{h\to0}\frac{\ln(x+h)-\ln(x)}{h}

Using the logarithm property, \ln(a)-\ln(b)=\ln\left(\frac{a}{b}\right)

We get, f'(x)=\lim_{h\to0}\frac{1}{h}\ln\left(\frac{x+h}{x}\right)

Since, \frac{x+h}{x}=1+\frac{h}{x}

We have, f'(x)=\lim_{h\to0}\frac{1}{h}\ln\left(1+\frac{h}{x}\right)

Let, u=\frac{h}{x}

Then, h=xu

As h→0, u→0.

Substituting h=xu, we get, f'(x)=\frac{1}{x}\lim_{u\to0}\frac{\ln(1+u)}{u}

Using the standard limit, \lim_{u\to0}\frac{\ln(1+u)}{u}=1

Therefore, f'(x)=\frac{1}{x}\times1=\frac{1}{x}

Hence, the derivative of ln(x) is:

\boxed{\frac{d}{dx}\ln(x)=\frac{1}{x}}

Example 1: Find the derivative of 3ln(x)?

Solution:

3ln(x)' = 3(1/x) = 3/x

Example 2: Find the derivative of ln(x)/5?

Solution:

(ln(x)/5)' = 1/5(ln(x))′ = (1/5) (1/x) = 1/5x