Computing derivatives and L'Hôpital's rule

Question 1: Derivative Rules (from file image_6640af.png)

The Question: Which of the following statements are correct? (Multiple Select Question)

Detailed Solution:

• Derivative of $e^{x^2} \cos x$: This requires the Product Rule and Chain Rule. Let $u=e^{x^2}$ and $v=\cos x$.

  • $u’ = e^{x^2} \cdot (2x)$ (Chain Rule)
  • $v’ = -\sin x$
  • $(uv)’ = u’v + uv’ = (2xe^{x^2})(\cos x) + (e^{x^2})(-\sin x) = e^{x^2}(2x \cos x - \sin x)$.
  • The first statement is TRUE.

• Derivative of $e^x \sin x$: This is a standard Product Rule application. Let $u=e^x$ and $v=\sin x$.

  • $u’=e^x$, $v’=\cos x$.
  • $(uv)’ = u’v+uv’ = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$.
  • The second statement is TRUE. The third statement is false.

• Derivative of $e^{x^2} \cos x$: (Repeated) The fourth statement provides a different incorrect answer.

Final Answer:

• Derivative of $e^{x^2} \cos x$ is $e^{x^2}(2x \cos x - \sin x)$.
• Derivative of $e^x \sin x$ is $e^x \sin x (\sin x + x \cos x)$. (Wait, the image says e^x sinx (sinx + xcosx). This must be a typo, it should be e^x(sinx+cosx). Let me check the image again. e^x sinx. The question is derivative of e^x sinx or e^(x sinx)? It’s e^x sinx. So my product rule is correct. The option in the image is wrong. Let me re-read all options in the image. Ah, the function is e^(x sinx).
• REVISED: Derivative of $e^{x\sin x}$: This is a chain rule problem. Let the exponent be $u = x \sin x$.
  • First, find $u’$ using the product rule: $u’ = (1)(\sin x) + (x)(\cos x) = \sin x + x\cos x$.
  • The derivative of $e^u$ is $e^u \cdot u’$.
  • So, the derivative is $e^{x\sin x}(\sin x + x\cos x)$. The second statement is TRUE.

Final Corrected Answer:

• Derivative of $e^{x^2} \cos x$ is $e^{x^2}(2x \cos x - \sin x)$.
• Derivative of $e^{x\sin x}$ is $e^{x\sin x}(\sin x + x\cos x)$.

Question 2: Trigonometric Limit (from file image_6640af.png)

The Question: $\lim_{x \to 0^+} \frac{\sin x}{\sqrt{x}}$ is __________.

Detailed Solution: This limit approaches the indeterminate form $\frac{0}{0}$. We can solve it by algebraic manipulation.

1. Rewrite the expression to isolate the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$. $$\frac{\sin x}{\sqrt{x}} = \frac{\sin x}{x} \cdot x \cdot \frac{1}{\sqrt{x}} = \frac{\sin x}{x} \cdot \sqrt{x}$$
2. Now, take the limit of the product: $$\lim_{x \to 0^+} \left( \frac{\sin x}{x} \cdot \sqrt{x} \right) = \left( \lim_{x \to 0^+} \frac{\sin x}{x} \right) \cdot \left( \lim_{x \to 0^+} \sqrt{x} \right)$$
3. Evaluate each limit: $$= (1) \cdot (0) = 0$$

Final Answer: The limit is 0.

Question 3: Derivative of $f(x) = x|x|$ (from file image_6640af.png)

The Question: Consider a function $f:\mathbb{R} \to \mathbb{R}$ such that $f(x)=x|x|$. Which of the following option(s) is(are) true?

Detailed Solution:

1. Rewrite as a piecewise function:
   • If $x \ge 0$, $|x|=x$, so $f(x) = x(x) = x^2$.
   • If $x < 0$, $|x|=-x$, so $f(x) = x(-x) = -x^2$. $$f(x) = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases}$$
2. Find the derivative for each piece:
   • For $x > 0$, the derivative of $x^2$ is $2x$.
   • For $x < 0$, the derivative of $-x^2$ is $-2x$.
3. Check differentiability at $x=0$:
   • Right-hand derivative: $\lim_{x \to 0^+} f’(x) = \lim_{x \to 0^+} 2x = 0$.
   • Left-hand derivative: $\lim_{x \to 0^-} f’(x) = \lim_{x \to 0^-} -2x = 0$.
   • Since the left and right derivatives are equal at $x=0$, the function is differentiable on $\mathbb{R}$.
4. Combine the derivative pieces:
   • The derivative is $2x$ for $x \ge 0$ and $-2x$ for $x < 0$. This can be written compactly as $f’(x) = 2|x|$. The options provide the piecewise definition.

Final Answer:

• $f$ is differentiable on $\mathbb{R}$.
• $f’(x) = \begin{cases} 2x & \text{if } x \ge 0 \ -2x & \text{if } x < 0 \end{cases}$

Question 4: Limit as a Derivative (from file image_664052.png)

The Question: If $f(x) = \sqrt{9-x^2}$, then find out the value of $\sqrt{8} \times \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1}$.

Detailed Solution:

1. Recognize the limit definition of the derivative: The expression $\lim_{x \to 1} \frac{f(x) - f(1)}{x - 1}$ is the definition of the derivative of $f(x)$ at the point $x=1$, which is $f’(1)$.
2. Find the derivative of $f(x)$:
   • Rewrite the function as $f(x) = (9-x^2)^{1/2}$.
   • Use the chain rule: $f’(x) = \frac{1}{2}(9-x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{9-x^2}}$.
3. Evaluate the derivative at $x=1$:
   • $f’(1) = \frac{-1}{\sqrt{9 - (1)^2}} = \frac{-1}{\sqrt{8}}$.
4. Calculate the final expression:
   • We need to find $\sqrt{8} \times f’(1)$.
   • $\sqrt{8} \times \left(\frac{-1}{\sqrt{8}}\right) = -1$.

Final Answer: The value is -1.

Question 5: Limit at Infinity (from file image_664052.png)

The Question: The value of $\lim_{x \to \infty} xe^{-x}$ is __________.

Detailed Solution:

1. Identify the Indeterminate Form: As $x \to \infty$, $x \to \infty$ and $e^{-x} \to 0$. This gives the indeterminate form $\infty \times 0$.
2. Rewrite as a Fraction: To use L’Hôpital’s rule, we must rewrite it as $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
   • $xe^{-x} = \frac{x}{e^x}$.
3. Check the New Form: As $x \to \infty$, the numerator $x \to \infty$ and the denominator $e^x \to \infty$. This is the form $\frac{\infty}{\infty}$.
4. Apply L’Hôpital’s Rule: Take the derivative of the numerator and the denominator separately.
   • $\lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{\frac{d}{dx}(x)}{\frac{d}{dx}(e^x)} = \lim_{x \to \infty} \frac{1}{e^x}$.
5. Evaluate the new limit: As $x \to \infty$, $e^x \to \infty$, so $\frac{1}{e^x} \to 0$.

Final Answer: The value is 0.

Question 6: Derivatives of Logarithms (from file image_664052.png)

The Question: Consider the following statements… Then the number of correct statements is __________.

Core Concept: The derivative of $\log_b u = \frac{u’}{u \ln b}$. For the natural log ($\ln$), since $\ln e = 1$, this simplifies to $\frac{u’}{u}$. To differentiate $\log_x 5$, we must first use the change of base formula: $\log_x 5 = \frac{\ln 5}{\ln x}$.

Detailed Solution:

• Statements 1 & 2 (Derivative of $\log_x 5$):
  • Let $f(x) = \log_x 5 = \frac{\ln 5}{\ln x} = (\ln 5)(\ln x)^{-1}$.
  • Use the chain rule: $f’(x) = (\ln 5) \cdot [-1 (\ln x)^{-2} \cdot \frac{1}{x}] = -\frac{\ln 5}{x(\ln x)^2}$.
  • We can rewrite this using $\ln x = \frac{\log_x e}{\log_x x} = \log_x e$. This doesn’t seem to simplify to the options.
  • Let’s try to rewrite the options. $-\frac{\log_x 5 \log_e e}{x^2}$. This is not correct.
  • Let’s use the provided answer as a guide: $-\frac{\log_x 5 \log_x e}{x}$. My calculation gives $-\frac{\ln 5}{x(\ln x)^2}$. Let’s convert my answer’s base. $\ln 5 = \frac{\log_x 5}{\log_x e}$ and $\ln x = \frac{1}{\log_x e}$.
  • $f’(x) = -\frac{(\frac{\log_x 5}{\log_x e})}{x(\frac{1}{\log_x e})^2} = -\frac{\log_x 5 \cdot \log_x e}{x}$. Statement 2 is correct, Statement 1 is incorrect.
• Statements 3 & 4 (Derivative of $\log_e x^2 = \ln(x^2)$):
  • Let $f(x) = \ln(x^2)$. Using the chain rule, $u=x^2, u’=2x$. The derivative is $\frac{u’}{u} = \frac{2x}{x^2} = \frac{2}{x}$.
  • Alternatively, simplify first: $\ln(x^2) = 2\ln x$. The derivative is $2(\frac{1}{x}) = \frac{2}{x}$.
  • Statement 4 is correct, Statement 3 is incorrect.

Final Answer: There are 2 correct statements.

Question 7: Applying L’Hôpital’s Rule (from file image_663ff4.png)

The Question: In which of the following, one can apply L’Hospital’s rule to evaluate the limits?

Detailed Solution: L’Hôpital’s rule can only be applied to limits that result in an indeterminate form of $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

• $\lim_{x \to \infty} \frac{x}{x+\sin x}$: As $x \to \infty$, $x \to \infty$. The term $\sin x$ oscillates between -1 and 1, so the denominator $x+\sin x$ also goes to $\infty$. This is of the form $\frac{\infty}{\infty}$. Can apply.
• $\lim_{x \to 0} \frac{\sin^2 x}{1-\cos(2x)}$: As $x \to 0$, $\sin^2 x \to 0$. As $x \to 0$, $\cos(2x) \to \cos(0) = 1$, so $1-\cos(2x) \to 0$. This is of the form $\frac{0}{0}$. Can apply.
• $\lim_{x \to \infty} \frac{7+\ln x}{x^3+6}$: As $x \to \infty$, $\ln x \to \infty$, so the numerator is $\infty$. The denominator $x^3+6 \to \infty$. This is of the form $\frac{\infty}{\infty}$. Can apply.
• $\lim_{x \to 0^+} \frac{1}{x+\sqrt{x}}$: As $x \to 0^+$, the denominator approaches 0. The numerator is 1. The result is $\frac{1}{0^+} \to +\infty$. This is not an indeterminate form. Cannot apply.

Final Answer: One can apply L’Hôpital’s rule to:

• $\lim_{x \to \infty} \frac{x}{x+\sin x}$
• $\lim_{x \to 0} \frac{\sin^2 x}{1-\cos(2x)}$
• $\lim_{x \to \infty} \frac{7+\ln x}{x^3+6}$

Question 8: Functional Equation (from file image_663ff4.png)

The Question: If $f(x+y) = f(x)f(y)$ for all $x,y \in \mathbb{R}$ and $f(9)=6, f’(0)=4$, then $f’(9)$ is __________.

Detailed Solution:

1. The functional equation $f(x+y)=f(x)f(y)$ implies that the function is an exponential function of the form $f(x) = a^x$.
2. Let’s use the definition of the derivative to find a general expression for $f’(x)$. $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
3. Using the given property, we can write $f(x+h) = f(x)f(h)$. $$f'(x) = \lim_{h \to 0} \frac{f(x)f(h)-f(x)}{h} = \lim_{h \to 0} \frac{f(x)(f(h)-1)}{h}$$
4. Since $f(x)$ does not depend on $h$, we can pull it out of the limit. $$f'(x) = f(x) \cdot \lim_{h \to 0} \frac{f(h)-1}{h}$$
5. Let’s look at the limit term. We need to find $f(0)$. From the original equation, $f(x) = f(x+0) = f(x)f(0)$. If we divide by $f(x)$ (assuming it’s not zero), we get $f(0)=1$.
6. The limit term is therefore $\lim_{h \to 0} \frac{f(h)-f(0)}{h}$, which is the definition of the derivative at $x=0$, i.e., $f’(0)$.
7. So we have a general relationship: $f’(x) = f(x) \cdot f’(0)$.
8. Now we can find $f’(9)$ using the given values.
   • $f’(9) = f(9) \cdot f’(0)$
   • $f’(9) = 6 \cdot 4 = 24$.

Final Answer: $f’(9)$ is 24.

Question 9: Properties of Differentiable Functions (from file image_663ff4.png)

The Question: Let $f$ and $g$ be two distinct functions from $\mathbb{R}$ to $\mathbb{R}$. Which of the following statements are true?

Detailed Solution:

• “If $fg$ is differentiable, then both $f$ and $g$ are differentiable.”: FALSE. A counterexample is $f(x)=|x|$ and $g(x)=|x|$. Neither is differentiable at $x=0$. But their product is $fg(x) = |x|^2 = x^2$, which is differentiable everywhere.
• “Assume that $g(x) \neq 0$… If $\frac{f}{g}$ is differentiable, then both $f$ and $g$ are differentiable.”: FALSE. The same counterexample works. Let $f(x)=|x|$ and $g(x)=|x|$ (for $x \neq 0$). Their quotient is $\frac{|x|}{|x|} = 1$, which is differentiable. But $f$ and $g$ are not differentiable at $x=0$.
• “If $f$ is an even differentiable function, then $f’$ is an odd function.”: TRUE. This is a standard property. If $f(-x) = f(x)$, we can differentiate both sides using the chain rule: $f’(-x) \cdot (-1) = f’(x) \implies -f’(-x) = f’(x) \implies f’(-x) = -f’(x)$, which is the definition of an odd function.
• “If $f$ is an odd differentiable function then, $f’$ is an even function.”: TRUE. This is also a standard property. If $f(-x) = -f(x)$, we differentiate both sides: $f’(-x) \cdot (-1) = -f’(x) \implies f’(-x) = f’(x)$, which is the definition of an even function.

Final Answer:

• If $f$ is an even differentiable function, then $f’$ is an odd function.
• If $f$ is an odd differentiable function then, $f’$ is an even function.

Question 10: Limit of a Trigonometric Product (from file image_663cf0.png)

The Question: The value of $\lim_{x \to 0^+} \cot(x) \sin(4x)$ is __________.

Detailed Solution:

1. Identify the Indeterminate Form: As $x \to 0^+$, $\cot(x) = \frac{\cos x}{\sin x} \to \frac{1}{0^+} \to \infty$. And $\sin(4x) \to 0$. This gives the indeterminate form $\infty \times 0$.
2. Rewrite as a Fraction: To solve, we rewrite $\cot(x)$ as $\frac{\cos x}{\sin x}$ to form a fraction suitable for standard limits or L’Hôpital’s rule. $$\lim_{x \to 0^+} \frac{\cos(x) \sin(4x)}{\sin(x)}$$
3. Solve using Standard Limits: This is now in the form $\frac{0}{0}$. We can rearrange it to use the limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$. $$\lim_{x \to 0^+} \cos(x) \cdot \frac{\sin(4x)}{x} \cdot \frac{x}{\sin(x)}$$
4. Group the limits: $$\left(\lim_{x \to 0^+} \cos(x)\right) \cdot \left(\lim_{x \to 0^+} \frac{\sin(4x)}{x}\right) \cdot \left(\lim_{x \to 0^+} \frac{x}{\sin(x)}\right)$$
5. Evaluate each limit:
   • $\lim_{x \to 0^+} \cos(x) = \cos(0) = 1$.
   • $\lim_{x \to 0^+} \frac{\sin(4x)}{x} = \lim_{x \to 0^+} 4 \cdot \frac{\sin(4x)}{4x} = 4 \cdot 1 = 4$.
   • $\lim_{x \to 0^+} \frac{x}{\sin(x)} = \frac{1}{\lim_{x \to 0^+} \frac{\sin(x)}{x}} = \frac{1}{1} = 1$.
6. Multiply the results: $$1 \times 4 \times 1 = 4$$

Final Answer: The value of the limit is 4.