Find from first principle the derivative of ( x) . [6] OR State Rolle's theorem. Interpret it geometrically. Discuss Rolle's theorem for the function f(x) = (x 1)(x 2) (x 3) in [1, 3]. Group 'B'

Model Answer: Find from first principle the derivative of $\log (\cos x)$

Step 1: Recall the definition of the derivative from first principles

The derivative of a function $f(x)$ from first principles is given by: $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

Here, $f(x)=\log (\cos x)$. So, we need to compute: $$\frac{d}{dx}\log (\cos x)=\underset{h\to 0}{\lim }\frac{\log (\cos (x+h))-\log (\cos x)}{h}$$

Step 2: Simplify the expression using logarithm properties

Using the property of logarithms, $\log a-\log b=\log \left(\frac{a}{b}\right)$, we rewrite the expression: $$\underset{h\to 0}{\lim }\frac{\log \left(\frac{\cos (x+h)}{\cos x}\right)}{h}$$

Step 3: Use the definition of the derivative of $\log u$

Recall that: $$\frac{d}{dx}\log u=\frac{1}{u}\cdot \frac{du}{dx}$$ Here, $u=\cos (x+h)$, but since we are taking the limit as $h\to 0$, we can think of this as a composition of functions. Instead, we directly apply the limit definition to $\log u$: $$\underset{h\to 0}{\lim }\frac{\log (u+\Delta u)-\log u}{h}=\frac{1}{u}\cdot \underset{h\to 0}{\lim }\frac{\Delta u}{h}$$ where $u=\cos x$ and $\Delta u=\cos (x+h)-\cos x$.

Thus, the expression becomes: $$\frac{1}{\cos x}\cdot \underset{h\to 0}{\lim }\frac{\cos (x+h)-\cos x}{h}$$

Step 4: Recognize the limit as the derivative of $\cos x$

The limit: $$\underset{h\to 0}{\lim }\frac{\cos (x+h)-\cos x}{h}$$ is the definition of the derivative of $\cos x$, which is $-\sin x$.

So, substituting back: $$\frac{d}{dx}\log (\cos x)=\frac{1}{\cos x}\cdot (-\sin x)=-\tan x$$

Final Answer

$$\boxed{-\tan x}$$

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Model Answer: State Rolle's Theorem, Interpret it Geometrically, and Discuss for $f(x)=(x-1)(x-2)(x-3)$ in [1, 3]

Part 1: Statement of Rolle's Theorem

Rolle's Theorem states that:

If a function $f$ satisfies the following conditions:

1. $f$ is continuous on the closed interval $[a,b]$,
2. $f$ is differentiable on the open interval $(a,b)$,
3. $f(a)=f(b)$,

then there exists at least one $c\in (a,b)$ such that $f^{\prime }(c)=0$.

Part 2: Geometric Interpretation

Geometrically, Rolle's Theorem implies that if a curve starts and ends at the same height (i.e., $f(a)=f(b)$), then there must be at least one point on the curve where the tangent is horizontal (i.e., the slope is zero). This means the curve has at least one local maximum or minimum (or a saddle point) between $a$ and $b$.

Part 3: Application to $f(x)=(x-1)(x-2)(x-3)$ on [1, 3]

Let $f(x)=(x-1)(x-2)(x-3)$.

Step 1: Check continuity on [1, 3] The function $f(x)$ is a polynomial, and polynomials are continuous everywhere, including on $[1,3]$. So, condition 1 is satisfied.

Step 2: Check differentiability on (1, 3) Polynomials are differentiable everywhere, so $f(x)$ is differentiable on $(1,3)$. Condition 2 is satisfied.

Step 3: Check $f(1)=f(3)$ Compute $f(1)$ and $f(3)$: $$f(1)=(1-1)(1-2)(1-3)=0\cdot (-1)\cdot (-2)=0$$ $$f(3)=(3-1)(3-2)(3-3)=2\cdot 1\cdot 0=0$$ Thus, $f(1)=f(3)$, so condition 3 is satisfied.

Conclusion from Rolle's Theorem: Since all three conditions are satisfied, there exists at least one $c\in (1,3)$ such that $f^{\prime }(c)=0$.

Step 4: Find the derivative and locate critical points First, expand $f(x)$: $$f(x)=(x-1)(x-2)(x-3)$$ Using the distributive property: $$f(x)=(x^2-3x+2)(x-3)=x^3-3x^2-3x^2+9x+2x-6=x^3-6x^2+11x-6$$ Now, compute the derivative: $$f^{\prime }(x)=3x^2-12x+11$$ Set $f^{\prime }(x)=0$: $$3x^2-12x+11=0$$ Solve the quadratic equation: $$x=\frac{12\pm \sqrt{(-12{)}^2-4\cdot 3\cdot 11}}{2\cdot 3}=\frac{12\pm \sqrt{144-132}}{6}=\frac{12\pm \sqrt{12}}{6}=\frac{12\pm 2\sqrt{3}}{6}=2\pm \frac{\sqrt{3}}{3}$$ Approximate the roots: $$x_1\approx 2-0.577=1.423,x_2\approx 2+0.577=2.577$$ Both $x_1$ and $x_2$ lie in the interval $(1,3)$, confirming Rolle's Theorem.

Final Answer

Rolle's Theorem guarantees that there exists at least one $c\in (1,3)$ such that $f^{\prime }(c)=0$. For $f(x)=(x-1)(x-2)(x-3)$, the derivative $f^{\prime }(x)=3x^2-12x+11$ has two critical points in $(1,3)$: $$\boxed{c=2\pm \frac{\sqrt{3}}{3}}$$