1. **Problem:** Differentiate $f(x) = x^2$ using the first principle of differentiation. 2. **Formula:** The first principle states: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Step:** Calculate $f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$ 4. **Step:** Substitute into the limit: $$f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h}$$ 5. **Step:** Simplify by canceling $h$: $$= \lim_{h \to 0} \frac{\cancel{h}(2x + h)}{\cancel{h}} = \lim_{h \to 0} (2x + h)$$ 6. **Step:** Evaluate the limit: $$f'(x) = 2x$$ 7. **Answer:** The derivative of $x^2$ by first principle is $2x$. --- 1. **Problem:** Verify Rolle's theorem for $f(x) = x^2 - 4x + 3$ on $[1,3]$. 2. **Conditions:** Rolle's theorem requires $f(a) = f(b)$ and $f$ continuous and differentiable on $[a,b]$. 3. **Step:** Calculate $f(1) = 1 - 4 + 3 = 0$ and $f(3) = 9 - 12 + 3 = 0$. 4. **Step:** Since $f(1) = f(3)$, conditions hold. 5. **Step:** Find $f'(x) = 2x - 4$. 6. **Step:** Solve $f'(c) = 0$ for $c$ in $(1,3)$: $$2c - 4 = 0 \Rightarrow c = 2$$ 7. **Answer:** Rolle's theorem holds with $c=2$. --- 1. **Problem:** State and explain the Mean Value Theorem (MVT). 2. **Statement:** If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ 3. **Explanation:** The slope of the tangent at $c$ equals the slope of the secant line joining $(a,f(a))$ and $(b,f(b))$. --- 1. **Problem:** Use Leibniz theorem to find the second derivative of $f(x) = x^3 \sin x$. 2. **Formula:** Leibniz theorem for product derivatives: $$\frac{d^n}{dx^n}(uv) = \sum_{k=0}^n \binom{n}{k} u^{(n-k)} v^{(k)}$$ 3. **Step:** For $n=2$, $$f''(x) = u''v + 2u'v' + uv''$$ 4. **Step:** Compute derivatives: $u = x^3$, $u' = 3x^2$, $u'' = 6x$ $v = \sin x$, $v' = \cos x$, $v'' = -\sin x$ 5. **Step:** Substitute: $$f''(x) = 6x \sin x + 2(3x^2)(\cos x) + x^3(-\sin x)$$ 6. **Step:** Simplify: $$f''(x) = 6x \sin x + 6x^2 \cos x - x^3 \sin x$$ 7. **Answer:** The second derivative is $$f''(x) = 6x \sin x + 6x^2 \cos x - x^3 \sin x$$ --- 1. **Problem:** Find the Maclaurin series expansion of $e^x$ up to $x^3$ term. 2. **Formula:** Maclaurin series: $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$$ 3. **Step:** For $f(x) = e^x$, all derivatives at 0 are 1. 4. **Step:** Write terms: $$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$ 5. **Answer:** Maclaurin series up to $x^3$ is $$1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$ --- 1. **Problem:** Evaluate the integral $$\int \frac{2x}{x^2 + 1} dx$$ using substitution. 2. **Step:** Let $u = x^2 + 1$, then $du = 2x dx$. 3. **Step:** Substitute: $$\int \frac{2x}{x^2 + 1} dx = \int \frac{1}{u} du$$ 4. **Step:** Integrate: $$= \ln|u| + C = \ln(x^2 + 1) + C$$ 5. **Answer:** The integral is $$\ln(x^2 + 1) + C$$ --- 1. **Problem:** Find the area under $y = x^2$ from $x=0$ to $x=2$. 2. **Formula:** Area $= \int_a^b f(x) dx$ 3. **Step:** Compute: $$\int_0^2 x^2 dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}$$ 4. **Answer:** Area is $\frac{8}{3}$ units squared. --- 1. **Problem:** Find the volume of revolution when $y = x$ is revolved about the x-axis from $x=0$ to $x=1$. 2. **Formula:** Volume $= \pi \int_a^b [f(x)]^2 dx$ 3. **Step:** Compute: $$\pi \int_0^1 x^2 dx = \pi \left[ \frac{x^3}{3} \right]_0^1 = \frac{\pi}{3}$$ 4. **Answer:** Volume is $\frac{\pi}{3}$ cubic units. --- 1. **Problem:** Find the equation of the plane passing through point $P(1,2,3)$ with normal vector $\vec{n} = (4,-5,6)$. 2. **Formula:** Plane equation: $$4(x - 1) - 5(y - 2) + 6(z - 3) = 0$$ 3. **Step:** Expand: $$4x - 4 - 5y + 10 + 6z - 18 = 0$$ 4. **Step:** Simplify: $$4x - 5y + 6z - 12 = 0$$ 5. **Answer:** Plane equation is $$4x - 5y + 6z = 12$$ --- 1. **Problem:** Find the dot product of vectors $\vec{a} = (1,2,3)$ and $\vec{b} = (4,-5,6)$. 2. **Formula:** Dot product: $$\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$$ 3. **Step:** Calculate: $$1 \times 4 + 2 \times (-5) + 3 \times 6 = 4 - 10 + 18 = 12$$ 4. **Answer:** Dot product is $12$.