Q.
Find, from the first principles, the derivatives of:
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First Principles Definition:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
i. Derivative of \( e^{\sqrt{x}} \):
\[
\begin{align}
f(x) &= e^{\sqrt{x}} \\
f'(x) &= \lim_{h \to 0} \frac{e^{\sqrt{x+h}} - e^{\sqrt{x}}}{h} \\
&Let, \sqrt{x} = y \text{ and } \sqrt{x+h} = y + k \\
&= \lim_{h, k \to 0} \frac{e^{y+k}-e^{y}}{h} \\
&= \lim_{h, k \to 0} \frac{e^y (e^k - 1)}{h} \\
&= e^{y} \lim_{h, k \to 0} \frac{e^{k} - 1}{k} \cdot \frac{k}{h} \\
&= e^{y} \lim_{h, k \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \\
&= e^{y} \lim_{h, k \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} \\
&= e^{y} \lim_{h, k \to 0} \frac{x+h - x}{h (\sqrt{x+h} + \sqrt{x})} \\
&= e^{y} \lim_{h, k \to 0} \frac{h}{h (\sqrt{x+h} + \sqrt{x})} \\
&= e^{y} \lim_{h, k \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} \\
&= \frac{e^{\sqrt{x}} }{2\sqrt{x}}
\end{align}
\]
ii. Derivative of \( e^{\sin x} \):
\[ \begin{align} f(x) &= e^{\sin x}, \\ f'(x) &= \lim_{h \to 0} \frac{e^{\sin(x+h)} - e^{\sin x}}{h}, \\ &= \lim_{h \to 0} \frac{e^{\sin x} \left( e^{\sin(x+h) - \sin x} - 1 \right)}{h}, \\ &= e^{\sin x} \lim_{h \to 0} \frac{e^{\sin(x+h) - \sin x} - 1}{h}. \end{align} \]
iii. Derivative of \( e^{\tan x} \):
\[ \begin{align} f(x) &= e^{\tan x}, \\ f'(x) &= \lim_{h \to 0} \frac{e^{\tan(x+h)} - e^{\tan x}}{h}, \\ &= \lim_{h \to 0} \frac{e^{\tan x} \left( e^{\tan(x+h) - \tan x} - 1 \right)}{h}, \\ &= e^{\tan x} \lim_{h \to 0} \frac{e^{\tan(x+h) - \tan x} - 1}{h}. \end{align} \]
iv. Derivative of \( e^{x^2} \):
\[ \begin{align} f(x) &= e^{x^2}, \\ f'(x) &= \lim_{h \to 0} \frac{e^{(x+h)^2} - e^{x^2}}{h}, \\ &= \lim_{h \to 0} \frac{e^{x^2} \left( e^{(x+h)^2 - x^2} - 1 \right)}{h}, \\ &= e^{x^2} \lim_{h \to 0} \frac{e^{(x+h)^2 - x^2} - 1}{h}. \end{align} \]
Q.
Find, from the first principles, the derivatives of:
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Q.
Find, from the first principles, the derivatives of:
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