Samacheer Kalvi Class 11 Business Maths Solutions Chapter 5 Differential Calculus Exercise 5.4

Question 1. Find the derivative of the following functions from first principle.
(i) \( x^2 \)
(ii) \( e^{-x} \)
(iii) \( \log(x+1) \)
Answer:
(i) To find the derivative of \( f(x) = x^2 \) from the first principle, we use the formula: \( \frac{d}{dx} f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).
First, find \( f(x+h) \): \( f(x+h) = (x+h)^2 \).
Next, substitute into the formula:
\( \frac{d}{dx} (x^2) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \)
\( = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2) - x^2}{h} \)
\( = \lim_{h \to 0} \frac{2xh + h^2}{h} \)
\( = \lim_{h \to 0} \frac{h(2x + h)}{h} \)
\( = \lim_{h \to 0} (2x + h) \)
Now, substitute \( h = 0 \):
\( = 2x + 0 \)
\( = 2x \)
So, the derivative of \( x^2 \) is \( 2x \). This method shows how the rate of change is calculated using small increments.

(ii) To find the derivative of \( f(x) = e^{-x} \) from the first principle, we use the formula: \( \frac{d}{dx} f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).
First, find \( f(x+h) \): \( f(x+h) = e^{-(x+h)} \).
Next, substitute into the formula:
\( \frac{d}{dx} (e^{-x}) = \lim_{h \to 0} \frac{e^{-(x+h)} - e^{-x}}{h} \)
\( = \lim_{h \to 0} \frac{e^{-x} e^{-h} - e^{-x}}{h} \)
\( = \lim_{h \to 0} \frac{e^{-x} (e^{-h} - 1)}{h} \)
\( = e^{-x} \lim_{h \to 0} \frac{e^{-h} - 1}{h} \)
We know that \( \lim_{k \to 0} \frac{e^k - 1}{k} = 1 \). Here, we have \( -h \) in the exponent, so we need to adjust the denominator:
\( = e^{-x} \lim_{h \to 0} \frac{e^{-h} - 1}{-h} \times (-1) \)
\( = e^{-x} (1) \times (-1) \)
\( = -e^{-x} \)
So, the derivative of \( e^{-x} \) is \( -e^{-x} \). The chain rule implicitly appears in this first principle derivation, as the derivative of \(-x\) is \(-1\).

(iii) To find the derivative of \( f(x) = \log(x+1) \) from the first principle, we use the formula: \( \frac{d}{dx} f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).
First, find \( f(x+h) \): \( f(x+h) = \log((x+1)+h) \).
Next, substitute into the formula:
\( \frac{d}{dx} (\log(x+1)) = \lim_{h \to 0} \frac{\log((x+1)+h) - \log(x+1)}{h} \)
Using the logarithm property \( \log a - \log b = \log (\frac{a}{b}) \):
\( = \lim_{h \to 0} \frac{1}{h} \log \left( \frac{x+1+h}{x+1} \right) \)
\( = \lim_{h \to 0} \frac{1}{h} \log \left( 1 + \frac{h}{x+1} \right) \)
We know that \( \lim_{k \to 0} \frac{\log(1+k)}{k} = 1 \). Here, \( k = \frac{h}{x+1} \). We need to adjust the expression to match this form:
\( = \lim_{h \to 0} \frac{1}{x+1} \times \frac{x+1}{h} \log \left( 1 + \frac{h}{x+1} \right) \)
\( = \frac{1}{x+1} \lim_{h \to 0} \frac{\log \left( 1 + \frac{h}{x+1} \right)}{\frac{h}{x+1}} \)
\( = \frac{1}{x+1} \times 1 \)
\( = \frac{1}{x+1} \)
So, the derivative of \( \log(x+1) \) is \( \frac{1}{x+1} \). This process shows the fundamental definition of a derivative in action.
In simple words: To find the derivative using the first principle, you calculate how much a function changes when its input changes by a tiny amount, and then you make that tiny change approach zero. For \( x^2 \), the derivative is \( 2x \). For \( e^{-x} \), it's \( -e^{-x} \). For \( \log(x+1) \), it's \( \frac{1}{x+1} \).

🎯 Exam Tip: Remember the two key limits used: \( \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \) and \( \lim_{h \to 0} \frac{\log(1+h)}{h} = 1 \). Adjust the expression carefully to match these forms.