From definition, find the derivative of e^ x . [6] OR State Mean value theorem. Verify it for the function f(x) = 2x^2 10x + 29 in [2, 7] .

Answer to the First Question: Find the derivative of $e^{\tan x}$ from definition

Step 1: Recall the definition of the derivative

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

Step 2: Apply the definition to $f(x)=e^{\tan x}$

Let $f(x)=e^{\tan x}$. Then, $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{e^{\tan (x+h)}-e^{\tan x}}{h}$$

Step 3: Factor out $e^{\tan x}$ and rewrite the expression

$$f^{\prime }(x)=e^{\tan x}\cdot \underset{h\to 0}{\lim }\frac{e^{\tan (x+h)-\tan x}-1}{h}$$

Step 4: Use the Taylor series expansion for $e^y$ around $y=0$

For small $h$, $\tan (x+h)-\tan x\approx h\cdot {\sec }^{2}x$ (since the derivative of $\tan x$ is ${\sec }^{2}x$). Thus, $$e^{\tan (x+h)-\tan x}\approx 1+(\tan (x+h)-\tan x)+\frac{(\tan (x+h)-\tan x{)}^2}{2}+\cdots$$ For small $h$, higher-order terms become negligible, so: $$e^{\tan (x+h)-\tan x}\approx 1+(\tan (x+h)-\tan x)$$

Step 5: Substitute back into the limit expression

$$f^{\prime }(x)=e^{\tan x}\cdot \underset{h\to 0}{\lim }\frac{1+(\tan (x+h)-\tan x)-1}{h}$$ $$=e^{\tan x}\cdot \underset{h\to 0}{\lim }\frac{\tan (x+h)-\tan x}{h}$$

Step 6: Recognize the derivative of $\tan x$

The limit ${\lim }_{h\to 0}\frac{\tan (x+h)-\tan x}{h}$ is the definition of the derivative of $\tan x$, which is ${\sec }^{2}x$.

Step 7: Final result

$$f^{\prime }(x)=e^{\tan x}\cdot {\sec }^{2}x$$

Final Answer: $$\boxed{e^{\tan x}{\sec }^{2}x}$$

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Answer to the Second Question: Mean Value Theorem (MVT) and Verification

Step 1: State the Mean Value Theorem

The Mean Value 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 $(a,b)$, then there exists at least one point $c\in (a,b)$ such that: $$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$$

Step 2: Verify MVT for $f(x)=2x^2-10x+29$ on $[2,7]$

Check Continuity and Differentiability

• The function $f(x)=2x^2-10x+29$ is a polynomial, so it is continuous and differentiable everywhere (including on $[2,7]$ and $(2,7)$). Thus, MVT applies.

Compute $f(2)$ and $f(7)$

$$f(2)=2(2{)}^2-10(2)+29=8-20+29=17$$ $$f(7)=2(7{)}^2-10(7)+29=98-70+29=57$$

Compute the slope of the secant line

$$\frac{f(7)-f(2)}{7-2}=\frac{57-17}{5}=\frac{40}{5}=8$$

Find the derivative $f^{\prime }(x)$

$$f^{\prime }(x)=\frac{d}{dx}(2x^2-10x+29)=4x-10$$

Set $f^{\prime }(c)=8$ and solve for $c$

$$4c-10=8$$ $$4c=18⟹c=\frac{18}{4}=4.5$$

Verify that $c\in (2,7)$

Since $2<4.5<7$, the point $c=4.5$ lies within the interval $(2,7)$.

Conclusion

The Mean Value Theorem is verified for $f(x)=2x^2-10x+29$ on $[2,7]$, with $c=4.5$ satisfying: $$f^{\prime }(4.5)=8=\frac{f(7)-f(2)}{7-2}$$

Final Answer: The Mean Value Theorem is verified with $c=\boxed{4.5}$.