CBSE Class 12 Mathematics Continuity And Differentiability Assignment Set 04

CHAPTER - 5

CONTINUITY AND DIFFERENTIATION

I. Derivative of an exponential function
Let \(f(x) = e^x\)
Then by definition,
\(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\)
\(= \lim_{h \to 0} \frac{e^{x + h} - e^x}{h}\)
\(= \lim_{h \to 0} \frac{e^x \cdot e^h - e^x}{h}\)
\(= \lim_{h \to 0} e^x \left(\frac{e^h - 1}{h}\right)\)
\(= e^x \left(\lim_{h \to 0} \frac{e^h - 1}{h}\right)\)
Since \(\lim_{h \to 0} \frac{e^h - 1}{h} = 1\), we get:
\(= e^x \cdot 1 = e^x\)
Hence, \(\frac{d}{dx}(e^x) = e^x\)
Similarly, we can write that \(\frac{d}{dx}(a^x) = a^x \log a\)
Answer: The derivative of the exponential function \(e^x\) with respect to \(x\) is derived above using the first principle of differentiation, yielding \(e^x\). For any general exponential function \(a^x\) where \(a > 0\) and \(a \neq 1\), the derivative is \(a^x \log a\).
In simple words: The derivative of \(e^x\) is equal to itself, while the derivative of \(a^x\) is \(a^x \log a\).

Exam Tip: Memorize the standard derivative formulas for exponential functions as they are frequently used directly in solving complex chain rule and logarithmic differentiation problems.

II. Derivative of a logarithmic function
Let \(f(x) = \log_e x\), where \(x > 0\)
Then, \(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\)
\(= \lim_{h \to 0} \frac{\log_e (x + h) - \log_e x}{h}\)
Using the logarithmic property \(\log a - \log b = \log\left(\frac{a}{b}\right)\):
\(= \lim_{h \to 0} \frac{\log_e \left(\frac{x + h}{x}\right)}{h}\)
\(= \lim_{h \to 0} \frac{\log_e \left(1 + \frac{h}{x}\right)}{h}\)
Multiply and divide the expression inside or rewrite to use the standard limit \(\lim_{u \to 0} \frac{\log(1 + u)}{u} = 1\):
\(= \lim_{h \to 0} \left[ \frac{\log_e \left(1 + \frac{h}{x}\right)}{\frac{h}{x}} \cdot \frac{1}{x} \right]\)
As \(h \to 0\), let \(u = \frac{h}{x}\), so \(u \to 0\):
\(= \frac{1}{x} \cdot \lim_{u \to 0} \frac{\log_e (1 + u)}{u} = \frac{1}{x} \cdot 1 = \frac{1}{x}\)
Hence, \(\frac{d}{dx}(\log_e x) = \frac{1}{x}\)
Answer: The derivative of the natural logarithmic function \(\log_e x\) with respect to \(x\) is proven using the first principle to be equal to \(\frac{1}{x}\) for \(x > 0\).
In simple words: The rate of change of the natural log of \(x\) is simply the reciprocal of \(x\).

Exam Tip: Always remember to state the domain restriction \(x > 0\) when dealing with logarithmic derivatives, since logarithms are undefined for zero and negative real numbers.