$$\\[5ex]$$ $$\text{if Point } A = (x,y) = (1,0), \text{ then } \cos(A) = 1 \\[2ex]$$ $$\\[5ex]$$ $$\text{if Point } A = (x,y) = (1,0), \text{ then } \sin(A) = 0 \\[2ex]$$ $$\\[5ex]$$ $$\text{if Point } B = (x,y) = (0,1), \text{ then } \sin(B) = 1 \\[2ex]$$ $$\\[5ex]$$ $$\text{if Point } B = (x,y) = (0,1), \text{ then } \cos(B) = 0 \\[2ex]$$ $$\\[5ex]$$

$$\\[5ex]$$ $$\text{If } \theta = 45^\circ = \frac{\pi}{4},\hspace{5pt} C = (0.707106811865, 0.707106811865)$$ $$\\[3ex]$$ $$\text{If } C = (0.707106811865, 0.707106811865),\hspace{5pt} C, sin = C.y = 0.707106811865$$ $$\\[3ex]$$ $$\text{If } C = (0.707106811865, 0.707106811865),\hspace{5pt} C, \cos = C.x = 0.707106811865$$ $$\\[5ex]$$ $$\text{If } \theta = 45^\circ = \frac{\pi}{4} ,\hspace{5pt} \sin(\theta) = 0.707106811865$$ $$\\[5ex]$$ $$\text{If } \theta = 45^\circ = \frac{\pi}{4},\hspace{5pt} \cos(\theta) = 0.707106811865$$ $$\\[5ex]$$ $$|\vec{C}| = \sqrt{x^2 + y^2}$$ $$\\[3ex]$$ $$\text{hipotenuse} = \hspace{3pt} h \hspace{3pt} = \sqrt{\text{base}^2 + \text{height}^2} \implies |\vec{C}| = \sqrt{x^2 + y^2}$$ $$\\[5ex]$$ $$|\vec{C}| = \sqrt{x^2 + y^2} = \hspace{3pt} 1 \hspace{3pt} = \sqrt{(\cos 45^\circ)^2 + (\sin 45^\circ)^2} = \sqrt{\left(\cos \left(\frac{\pi}{4}\right)\right)^2 + \left(\sin \left(\frac{\pi}{4}\right)\right)^2}$$

$$\LARGE\text{Identity \#1}$$

$$\\[5ex]$$ $$\boxed{\quad\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)\quad} \\[5ex]$$ $$\\[5ex]$$ $$\text{Point } P_0: \text{At angle } 0 \implies (x, y) = (1, 0)$$ $$\\[5ex]$$ $$\text{Point } P_1: \text{At angle } B \implies (x, y) = (\cos B, \sin B)$$ $$\\[5ex]$$ $$\text{Point } P_2: \text{At angle } -A \implies (x, y) = (\cos(-A), \sin(-A)), \text{ which simplifies to } (\cos A, -\sin A)$$ $$\\[5ex]$$ $$\text{Point } P_3: \text{At angle } A+B \implies (x, y) = (\cos(A+B), \sin(A+B))$$

$$\\[5ex]$$ $$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \quad \text{(derived from } a^2 + b^2 = c^2\text{)}$$ $$\\[5ex]$$ $$(d_{P_1 P_2})^2 = (d_{P_0 P_3})^2$$ $$\\[5ex]$$ $$\text{For } P_0 \text{ to } P_3:$$ $$\\[5ex]$$ $$(d_{P_0 P_3})^2 = (\cos(A+B) - 1)^2 + (\sin(A+B) - 0)^2$$ $$\\[5ex]$$ $$(d_{P_0 P_3})^2 = \cos^2(A+B) - 2\cos(A+B) + 1 + \sin^2(A+B)$$ $$\\[5ex]$$ $$\sin^2\theta + \cos^2\theta = 1$$ $$\\[5ex]$$ $$(d_{P_0 P_3})^2 = \underbrace{\cos^2(A+B) + \sin^2(A+B)}_{1} - 2\cos(A+B) + 1$$ $$\\[5ex]$$ $$(d_{P_0 P_3})^2 = 1 - 2\cos(A+B) + 1$$ $$\\[5ex]$$ $$(d_{P_0 P_3})^2 = 2 - 2\cos(A + B)$$ $$\\[5ex]$$ $$\text{For } P_1 \text{ to } P_2:$$ $$\\[5ex]$$ $$(d_{P_1 P_2})^2 = (\cos B - \cos A)^2 + (\sin B - (-\sin A))^2$$ $$\\[5ex]$$ $$(d_{P_1 P_2})^2 = (\cos B - \cos A)^2 + (\sin B + \sin A)^2$$ $$\\[5ex]$$ $$(d_{P_1 P_2})^2 = (\cos^2 B - 2\cos A \cos B + \cos^2 A) + (\sin^2 B + 2\sin A \sin B + \sin^2 A)$$ $$\\[5ex]$$ $$(\cos^2 B + \sin^2 B = 1) \text{ and } (\cos^2 A + \sin^2 A = 1)$$ $$\\[5ex]$$ $$(d_{P_1 P_2})^2 = \underbrace{\cos^2 B + \sin^2 B}_{1} + \underbrace{\cos^2 A + \sin^2 A}_{1} - 2\cos A \cos B + 2\sin A \sin B$$ $$\\[5ex]$$ $$(d_{P_1 P_2})^2 = 1 + 1 - 2\cos A \cos B + 2\sin A \sin B$$ $$\\[5ex]$$ $$(d_{P_1 P_2})^2 = 2 - 2\cos A \cos B + 2\sin A \sin B$$ $$\\[5ex]$$ $$(d_{P_0 P_3})^2 = (d_{P_1 P_2})^2 $$ $$\\[5ex]$$ $$2 - 2\cos(A+B) = 2 - 2\cos A \cos B + 2\sin A \sin B$$ $$\\[5ex]$$ $$-2\cos(A+B) = -2\cos A \cos B + 2\sin A \sin B$$ $$\\[5ex]$$ $$\large \boxed {\quad \cos(A+B) = \cos A \cos B - \sin A \sin B \quad }$$ $$\\[5ex]$$

$$\LARGE\text{Identity \#2}$$

$$\\[5ex]$$ $$\boxed{\large \quad \mathbf{\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)}\quad}$$

$$\begin{aligned} &\text{1. The Geometric Setup} \\ &\text{Imagine two angles, } \alpha \text{ (which we'll call } x\text{) and } \beta \text{ (which we'll call } h\text{), stacked on top of each other:} \\ &\text{Draw angle } x \text{ from the } x\text{-axis.} \\ &\text{Draw angle } h \text{ starting from the terminal side of } x\text{.} \\ &\text{The total angle from the } x\text{-axis is } (x + h)\text{.} \\ &\text{We place a point } P \text{ on the terminal side of the total angle such that the distance from the origin } O \\ &\text{ to } P \text{ is 1 unit } (OP = 1)\text{.} \\[5ex] &\text{2. Identifying the Components} \\ &\text{To find } \sin(x + h)\text{, we need the vertical height of point } \\ &P \text{ relative to the } x\text{-axis. Let’s drop some perpendicular lines to create right triangles:} \\ &\text{Drop a perpendicular from } P \text{ to the terminal side of angle } x \text{ at point } Q\text{.} \\ &\text{Drop a perpendicular from } P \text{ to the } x\text{-axis at point } R\text{. } PR \text{ is our goal, because } PR = \sin(x+h)\text{.} \\ &\text{Drop a perpendicular from } Q \text{ to the } x\text{-axis at point } S\text{.} \\ &\text{Draw a horizontal line from } Q \text{ to the line } PR\text{, hitting it at point } T\text{.} \\ &\text{Now, notice that the total height } PR \text{ is composed of two segments:} \\ &PR = PT + TR \\ &\text{Since } TRQS \text{ is a rectangle, } TR = QS\text{. So:} \\ &\sin(x+h) = PT + QS \end{aligned}$$

$$\\[5ex]$$ $$\begin{aligned} &\text{3. Solving for the Segments} \\ &\text{We now look at the two right triangles we've created:} \\ &\text{Triangle 1: } \triangle OQP \\ &\text{The hypotenuse } OP = 1\text{.} \\ &PQ = \sin(h) \\ &OQ = \cos(h) \\ &\text{Triangle 2: } \triangle OSQ \\ &\text{This triangle has angle } x \text{ and hypotenuse } OQ \text{ (which we know is } \cos(h)\text{).} \\ &\sin(x) = \frac{QS}{OQ} \implies QS = OQ \sin(x) \\ &\text{Substitute } OQ\text{: } QS = \cos(h) \sin(x) \\ &\text{Triangle 3: } \triangle TQP \\ &\text{This is the clever part. The angle } \angle TPQ \text{ is actually equal to } x \\ &\text{ (because its sides are perpendicular to the sides of angle } x\text{).} \\ &\cos(x) = \frac{PT}{PQ} \implies PT = PQ \cos(x) \\ &\text{Substitute } PQ\text{: } PT = \sin(h) \cos(x) \end{aligned}$$ $$\\[5ex]$$

$$\begin{aligned} &\text{4. Combining the Results} \\ &\text{Now we plug our values for } QS \text{ and } PT \text{ back into our original height equation:} \\ &\sin(x+h) = PT + QS \\ &\sin(x+h) = (\sin(h) \cos(x)) + (\cos(h) \sin(x)) \\ &\text{Rearranging for the standard form:} \\ &\mathbf{\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)} \end{aligned}$$

$$\\[5ex]$$ $$\boxed{\large \quad \mathbf{\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)}\quad}$$

$$\LARGE\text{Limits \#1}$$

$$\\[5ex]$$ $$\large\boxed {\lim_{h \to 0} \frac{\sin(h)}{h} = 1}$$ $$\\[5ex]$$ $$\begin{aligned} &\text{\textbf{Geometric Setup}} \\ &\text{Consider a unit circle with radius } r = 1 \text{ and an angle } h \text{ where } 0 < h < \frac{\pi}{2}. \\ &\text{We define three regions based on this angle:} \\ \\ &\text{1. \textbf{Small Triangle:}} \text{ vertices at } (0,0), (1,0), \text{ and } (\cos(h), \sin(h)). \\ &\text{Area}_{\text{triangle 1}} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 1 \cdot \sin(h) = \frac{1}{2}\sin(h) \\ \\ &\text{2. \textbf{Circular Sector:}} \text{ with central angle } h. \\ &\text{Area}_{\text{sector}} = \frac{1}{2} r^2 h = \frac{1}{2} (1)^2 h = \frac{1}{2}h \\ \\ &\text{3. \textbf{Large Triangle:}} \text{ formed by the base and the tangent line at } (1,0). \\ &\text{Area}_{\text{triangle 2}} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 1 \cdot \tan(h) = \frac{1}{2}\tan(h) \\[7ex] &\textbf{Establishing the Inequality} \\ \\ &\text{From the geometry of the unit circle, it is clear that:} \\[8px] &\text{Area}_{\text{triangle 1}} < \text{Area}_{\text{sector}} < \text{Area}_{\text{triangle 2}} \\[8px] &\text{Substituting the formulas derived above:} \\[8px] &\frac{1}{2}\sin(h) < \frac{1}{2}h < \frac{1}{2}\tan(h) \\[8px] &\text{Multiply the entire inequality by } 2 \text{ to simplify:} \\[5px] &\sin(h) < h < \tan(h) \\[7ex] &\text{\textbf{3. Algebraic Manipulation}} \\ &\text{Since we are interested in } \frac{\sin(h)}{h}, \text{ we divide the inequality by } \sin(h). \\ &\text{Because } h \text{ is in the first quadrant } (0 < h < \frac{\pi}{2}), \sin(h) \text{ is positive, so the} \\ &\text{inequality signs do not flip:} \\ \\ &\frac{\sin(h)}{\sin(h)} < \frac{h}{\sin(h)} < \frac{\tan(h)}{\sin(h)} \\ \\ &\text{Recalling that } \tan(h) = \frac{\sin(h)}{\cos(h)}, \text{ we get:} \\ \\ &1 < \frac{h}{\sin(h)} < \frac{1}{\cos(h)} \\[7ex] &\textbf{Reciprocal Rule:} \\ &\text{When you take the reciprocal of all parts of an inequality where all} \\ &\text{terms are positive, the inequality symbols must flip.} \\[1em] &2 < 3 < 4 \\[0.5em] &\frac{1}{2} > \frac{1}{3} > \frac{1}{4} \\[7ex] \end{aligned}$$

$$\begin{aligned} \boxed { \text{The reciprocal of a function is simply } 1 \text{ divided by that function.} \\ \text{If you have a function } f(x), \text{ its reciprocal is:} \\[1em] \quad \frac{1}{f(x)} \quad \text{or} \quad [f(x)]^{-1} } \end{aligned}$$ $$\\[5ex]$$

$$\begin{aligned} &\text{Now, take the reciprocal of all parts. Taking the reciprocal of} \\ &\text{positive terms reverses the inequality signs:} \\ \\ &1 > \frac{\sin(h)}{h} > \cos(h) \\ \\ &\text{Or, written more standardly:} \\ \\ &\cos(h) < \frac{\sin(h)}{h} < 1\\ \\ &\ {\lim_{h \to 0}cos(h) < \lim_{h \to 0}\frac{\sin(h)}{h} < \lim_{h \to 0} 1}\\ \\ &\text{We now take the limit as } h \to 0 \text{ for the outer functions:} \\ \\ &\lim_{h \to 0} 1 = 1 \\ &\lim_{h \to 0} \cos(h) = \cos(0) = 1 \\ \\ &1 \le \lim_{h \to 0} \frac{\sin(h)}{h} \le 1 \\[5ex] &\text{Since } \frac{\sin(h)}{h} \text{ is "squeezed" between two functions that both approach } 1 \\ &\text{as } h \text{ approaches } 0, \text{ the Squeeze Theorem states that:} \\ \\ &\boxed{\color{white}{\lim_{h \to 0^+} \frac{\sin(h)}{h} = 1}} \end{aligned}$$ $$\\[5ex]$$

$$\LARGE\text{Limits \#2}$$

$$\boxed{\color{white}{\lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0}}$$ $$\\[5ex]$$ $$\lim_{h \to 0} \frac{\cos(h) - 1}{h} \cdot \frac{\cos(h) + 1}{\cos(h) + 1}$$ $$\\[5ex]$$ $$\lim_{h \to 0} \frac{\cos^2(h) - 1}{h(\cos(h) + 1)}$$ $$\\[5ex]$$ $$\sin^2(h) + \cos^2(h) = 1 \implies \cos^2(h) - 1 = -\sin^2(h)$$ $$\\[5ex]$$ $$\lim_{h \to 0} \frac{-\sin^2(h)}{h(\cos(h) + 1)}$$ $$\\[5ex]$$ $$\lim_{h \to 0} \left( \frac{\sin(h)}{h} \cdot \frac{-\sin(h)}{\cos(h) + 1} \right)$$ $$\\[5ex]$$ $$\lim_{x \to a} [f(x) \cdot g(x)] = \left( \lim_{x \to a} f(x) \right) \cdot \left( \lim_{x \to a} g(x) \right)$$ $$\\[5ex]$$ $$\text{if, }\quad \lim_{h \to 0} \frac{\sin(h)}{h} = 1$$ $$\\[5ex]$$ $$\text{if, } h = 0 \implies \frac{-\sin(0)}{\cos(0) + 1} = \frac{0}{1 + 1} = \frac{0}{2} = 0$$ $$\\[5ex]$$ $$(1) \cdot (0) = 0$$ $$\\[5ex]$$ $$\boxed{\lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0}$$ $$\\[5ex]$$

$$\LARGE\text{Limits \#3}$$

$$\large\boxed{\lim_{h \to 0} \frac{e^h - 1}{h} = 1}$$ $$\\[5ex]$$ $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$$ $$\\[5ex]$$ $$\text{if, } h = x,\quad e^h = 1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + \dots$$ $$\\[5ex]$$ $$\lim_{h \to 0} \frac{(1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + \dots) - 1}{h}$$ $$\\[5ex]$$ $$\lim_{h \to 0} \frac{h + \frac{h^2}{2!} + \frac{h^3}{3!} + \dots}{h}$$ $$\\[5ex]$$ $$\lim_{h \to 0} \left( \frac{h}{h} + \frac{h^2}{2!h} + \frac{h^3}{3!h} + \dots \right)$$ $$\\[5ex]$$ $$\lim_{h \to 0} \left( \frac{h}{h} + \frac{h^2}{2!h} + \frac{h^3}{3!h} + \dots \right)$$ $$\\[5ex]$$ $$\lim_{h \to 0} \left( 1 + \frac{h}{2!} + \frac{h^2}{3!} + \dots \right)$$ $$\\[5ex]$$ $$1 + 0 + 0 + \dots = 1$$ $$\\[5ex]$$ $$\boxed{\lim_{h \to 0} \frac{e^h - 1}{h} = 1} $$

$$\text{General Formula for Derivative}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$ $$\\[5ex]$$

$$\LARGE\text{Example \#1}$$

$$\\[5ex]$$ $$\large \boxed {\quad f(x) = x^2 \quad}$$ $$\\[5ex]$$ $$f(x+h) = (x+h)^2$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$$ $$\\[5ex]$$ $$(x+h)^2 = x^2 + 2xh + h^2$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{\color{red}{\cancel{x^2}} \color{white}{+ 2xh + h^2} \color{red}{\cancel{- x^2}}}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{\color{red}{\cancel{h}} \color{white}{(2x + h)}}{\color{red}{\cancel{h}}}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} (2x + h)$$ $$\\[5ex]$$ $$f'(x) = 2x + 0$$ $$\\[5ex]$$ $$\large \boxed {\quad f'(x) = 2x \quad}$$ $$\\[5ex]$$

$$\large \text{Theorem \#1 }$$

$$\\[5ex]$$ $$\large \boxed {\quad f(x) = x^n \quad}$$ $$\\[5ex]$$ $$f(x) = x^n \implies f'(x) = nx^{n-1}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{(x + h)^n - x^n}{h}$$ $$\\[5ex]$$ $$\text{Binomial Theorem}$$ $$(x + h)^n = x^n + nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \dots + h^n$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{\left( \color{red}{\cancel{x^n}} \color{white}{+ nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \dots + h^n} \right) \color{red}{\cancel{- x^n}}}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \dots + h^n}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{h \left( nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \dots + h^{n-1} \right)}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{\color{red}{\cancel{h}} \color{white}{\left( nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \dots + h^{n-1} \right)}}{\color{red}{\cancel{h}}}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \left( nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \dots + h^{n-1} \right)$$ $$\\[5ex]$$ $$f'(x) = nx^{n-1} + 0 + 0 + \dots + 0$$ $$\\[5ex]$$ $$\large \boxed {\quad f'(x) = nx^{n-1}\quad}$$ $$\\[5ex]$$

$$\large \text{Theorem \#2 }$$

$$\\[5ex]$$ $${\quad f(x) = \cos(x_1) \quad}$$ $$\\[5ex]$$ $$\large \boxed {f'(x) = \lim_{h \to 0} \frac{\cos(x + h) - \cos(x)}{h}}$$ $$\\[5ex]$$ $$\cos(x + h) = \cos(x)\cos(h) - \sin(x)\sin(h)$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{\cos(x)\cos(h) - \cos(x) - \sin(x)\sin(h)}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \left[ \cos(x) \left( \frac{\cos(h) - 1}{h} \right) - \sin(x) \left( \frac{\sin(h)}{h} \right) \right]$$ $$\\[5ex]$$ $$\text{if, } \quad \lim_{h \to 0} \frac{\sin(h)}{h} = 1$$ $$\\[5ex]$$ $$\text{if, }\quad \lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0$$ $$\\[5ex]$$ $$f'(x) = \cos(x) \cdot (0) - \sin(x) \cdot (1)$$ $$\\[5ex]$$ $$\boxed{f'(x) = -\sin(x)}$$ $$\\[5ex]$$

$$\large \text{Theorem \#3 }$$

$$\\[5ex]$$ $$\large \boxed {f'(x) = \lim_{h \to 0} \frac{\sin(x + h) - \sin(x)}{h}}$$ $$\\[5ex]$$ $$\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \left[ \frac{\sin(x)(\cos(h) - 1)}{h} + \frac{\cos(x)\sin(h)}{h} \right]$$ $$\\[5ex]$$ $$f'(x) = \sin(x) \cdot \lim_{h \to 0} \left( \frac{\cos(h) - 1}{h} \right) + \cos(x) \cdot \lim_{h \to 0} \left( \frac{\sin(h)}{h} \right)$$ $$\\[5ex]$$ $$\lim_{h \to 0} \frac{\sin(h)}{h} = 1$$ $$\\[5ex]$$ $$\lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0$$ $$\\[5ex]$$ $$f'(x) = \sin(x) \cdot (0) + \cos(x) \cdot (1)$$ $$\\[5ex]$$ $$f'(x) = \cos(x)$$ $$\\[5ex]$$

$$\large \text{Theorem \#4 }$$

$$\\[5ex]$$ $$\large \boxed {f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h}$$ $$\\[5ex]$$ $$e^{a+b} = e^{a} \cdot e^{b},\quad e^{x+h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{e^x \cdot e^h - e^x}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{e^x(e^h - 1)}{h}$$ $$\\[5ex]$$ $$f'(x) = e^x \cdot \left[ \lim_{h \to 0} \frac{e^h - 1}{h} \right]$$ $$\\[5ex]$$ $$\lim_{h \to 0} \frac{e^h - 1}{h} = 1$$ $$\\[5ex]$$ $$f'(x) = e^x \cdot (1)$$ $$\\[5ex]$$ $$\large \boxed {f'(x) = e^x}$$ $$\\[5ex]$$