From definition, find the derivative of ^ 1x . [6] OR Using Mean Value Theorem, find a point on the curve f(x) = x^2 2x , the tangent at which is parallel to the chord joining the points (1, 1) and…

Model Answer 1: Find the derivative of $\log {\cos }^{-1}x$ from definition

Step 1: Recall the definition of the derivative

The derivative of a function $f(x)$ is defined as: $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$ Here, $f(x)=\log {\cos }^{-1}x$.

Step 2: Write the derivative using the limit definition

$$\frac{d}{dx}\left(\log {\cos }^{-1}x\right)=\underset{h\to 0}{\lim }\frac{\log {\cos }^{-1}(x+h)-\log {\cos }^{-1}x}{h}$$

Step 3: Simplify the expression using logarithm properties

$$=\underset{h\to 0}{\lim }\frac{1}{h}\log \left(\frac{{\cos }^{-1}(x+h)}{{\cos }^{-1}x}\right)$$

Step 4: Use the Taylor series expansion for ${\cos }^{-1}(x+h)$ around $h=0$

The first-order Taylor expansion of ${\cos }^{-1}(x+h)$ is: $${\cos }^{-1}(x+h)\approx {\cos }^{-1}x-\frac{h}{\sqrt{1-x^2}}+O(h^2)$$ Thus, $$\frac{{\cos }^{-1}(x+h)}{{\cos }^{-1}x}\approx 1-\frac{h}{\sqrt{1-x^2}{\cos }^{-1}x}$$

Step 5: Substitute back into the limit expression

$$=\underset{h\to 0}{\lim }\frac{1}{h}\log \left(1-\frac{h}{\sqrt{1-x^2}{\cos }^{-1}x}\right)$$

Step 6: Use the approximation $\log (1+y)\approx y$ for small $y$

$$=\underset{h\to 0}{\lim }\frac{1}{h}\left(-\frac{h}{\sqrt{1-x^2}{\cos }^{-1}x}\right)=-\frac{1}{\sqrt{1-x^2}{\cos }^{-1}x}$$

Step 7: Final simplification

$$\frac{d}{dx}\left(\log {\cos }^{-1}x\right)=-\frac{1}{\sqrt{1-x^2}{\cos }^{-1}x}$$

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Model Answer 2: Using Mean Value Theorem (MVT) to find a point on $f(x)=x^2-2x$

Step 1: Recall the Mean Value Theorem

The MVT states that if a function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a point $c\in (a,b)$ such that: $$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$$

Step 2: Identify the given points and compute the slope of the chord

Given points: $A(1,-1)$ and $B(4,8)$.

Compute $f(1)$ and $f(4)$: $$f(1)=(1{)}^2-2(1)=1-2=-1$$ $$f(4)=(4{)}^2-2(4)=16-8=8$$

Slope of the chord $AB$: $$m=\frac{f(4)-f(1)}{4-1}=\frac{8-(-1)}{3}=\frac{9}{3}=3$$

Step 3: Find the derivative of $f(x)$

$$f(x)=x^2-2x⟹f^{\prime }(x)=2x-2$$

Step 4: Apply MVT to find $c$

Set $f^{\prime }(c)=3$: $$2c-2=3⟹2c=5⟹c=\frac{5}{2}=2.5$$

Step 5: Verify that $c$ lies in $(1,4)$

Since $1<2.5<4$, the point $c=2.5$ satisfies the MVT condition.

Step 6: Find the tangent point

The tangent at $x=2.5$ is parallel to the chord $AB$. The point on the curve is: $$f(2.5)=(2.5{)}^2-2(2.5)=6.25-5=1.25$$ Thus, the point is $(2.5,1.25)$.

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Final Answer (for the OR part)

The point on the curve $f(x)=x^2-2x$ where the tangent is parallel to the chord joining $(1,-1)$ and $(4,8)$ is (2.5, 1.25).