Differentiation Rules - IB Mathematics Analysis and Approaches

1. Let $f(x) = x^2 e^{3x}$. Find $f'(x)$.
   Answer: $f'(x) = e^{3x}(2x + 3x^2)$ To find the derivative of $f(x) = x^2 e^{3x}$, we must use the product rule, $(uv)' = u'v + uv'$, where $u = x^2$ and $v = e^{3x}$. First, find the derivatives of $u$ and $v$: $u' = 2x$ To find $v'$, we need the chain rule. The derivative of $e^{3x}$ is $e^{3x} \cdot 3 = 3e^{3x}$. Now, apply the product rule: $f'(x) = (2x)(e^{3x}) + (x^2)(3e^{3x})$ Finally, factor out the common term $xe^{3x}$ for the simplified form, but the answer choices have $e^{3x}$ factored out. $f'(x) = e^{3x}(2x + 3x^2)$ * **A: $f'(x) = 6x e^{3x}$** is incorrect. This results from incorrectly multiplying the derivatives of the two parts, i.e., $(2x)(3e^{3x})$. * **B: $f'(x) = e^{3x}(2x + x^2)$** is incorrect. This happens if the chain rule is forgotten when differentiating $e^{3x}$, treating its derivative as just $e^{3x}$. * **C: $f'(x) = e^{3x}(2x + 3x^2)$** is correct. * **D: $f'(x) = e^{3x}(2 + 3x)$** is incorrect. This would arise from incorrectly factoring $x$ out of $2x + 3x^2$.
2. Find the x-coordinates of the stationary points of the function $f(x) = x^3 e^{-x}$.
   Answer: x = 0 and x = 3 To find the stationary points, we need to find the derivative $f'(x)$ and set it to zero. Use the product rule with $u = x^3$ and $v = e^{-x}$. $u' = 3x^2$ $v' = -e^{-x}$ (using the chain rule) $f'(x) = u'v + uv' = (3x^2)(e^{-x}) + (x^3)(-e^{-x})$ Now, factor the expression: $f'(x) = e^{-x}(3x^2 - x^3) = x^2e^{-x}(3-x)$ Set $f'(x) = 0$: $x^2e^{-x}(3-x) = 0$ Since $e^{-x}$ is never zero, we have two possibilities for the product to be zero: 1. $x^2 = 0 \implies x = 0$ 2. $3-x = 0 \implies x = 3$ So, the x-coordinates of the stationary points are 0 and 3. * A: x = 3 only is incorrect. This ignores the solution from the $x^2$ factor. * B: x = 0 only is incorrect. This ignores the solution from the $(3-x)$ factor. * C: x = 0 and x = 3 is correct. * D: x = 0 and x = -3 is incorrect. This comes from a sign error when solving $3-x=0$.
3. Let $f(x) = x \ln x$ for $x>0$. Find $f''(x)$.
   Answer: $\frac{1}{x}$ To find the second derivative, $f''(x)$, we must first find the first derivative, $f'(x)$. Using the product rule on $f(x) = x \ln x$ with $u=x$ and $v=\ln x$: $u'=1$, $v'=1/x$ $f'(x) = u'v + uv' = (1)(\ln x) + (x)(\frac{1}{x}) = \ln x + 1$ Now, differentiate $f'(x)$ to find $f''(x)$: $f''(x) = \frac{d}{dx}(\ln x + 1) = \frac{1}{x} + 0 = \frac{1}{x}$ * **A: $-\frac{1}{x^2}$** is incorrect. This would be the second derivative if the product rule was ignored and $f'(x)$ was mistakenly taken as just $1/x$. * **B: $\frac{1}{x}$** is correct. * C: 1 is incorrect. This may result from incorrectly differentiating $\ln x$ as 1. * **D: $\ln x + 1$** is incorrect. This is the first derivative, $f'(x)$, not the second.
4. The tangent to the graph of $f(x) = ae^x - 3x$ at $x=0$ is parallel to the line $y=2x+1$. Find the value of $a$.
   Answer: 5 The gradient of the line $y=2x+1$ is 2. Since the tangent to the curve is parallel to this line, the gradient of the tangent at $x=0$ must also be 2. The gradient of the tangent is given by the derivative $f'(x)$. First, find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(ae^x - 3x) = ae^x - 3$ Now, evaluate the derivative at $x=0$ and set it equal to the required gradient, 2: $f'(0) = ae^0 - 3 = 2$ Since $e^0 = 1$: $a(1) - 3 = 2$ $a - 3 = 2$ $a = 5$ * A: 1 is incorrect. This value is obtained by incorrectly setting the function value $f(0)$ equal to the y-intercept of the line, 1. So $ae^0 - 3(0) = 1 \implies a=1$. * B: 2 is incorrect. This value is obtained by incorrectly setting the function value $f(0)$ equal to the gradient, 2. So $ae^0 - 3(0) = 2 \implies a=2$. * C: 4 is incorrect. This could be obtained by incorrectly setting $f'(x) = 2x+1$ and evaluating at $x=0$: $ae^0 - 3 = 2(0) + 1 \implies a-3 = 1 \implies a=4$. * D: 5 is correct.
5. Given $y = e^{\cos^2 x}$, find $\frac{dy}{dx}$.
   Answer: $-\sin(2x) e^{\cos^2 x}$ This requires the chain rule. The function can be seen as $y = e^u$ where $u = v^2$ and $v = \cos x$. 1. Derivative of the outer function $e^u$ is $e^u$, so we have $e^{\cos^2 x}$. 2. Derivative of the middle function $u = v^2 = (\cos x)^2$ is $2v \cdot v' = 2\cos x \cdot (-\sin x)$. Combining these using the chain rule: $\frac{dy}{dx} = e^{\cos^2 x} \cdot (2\cos x)(-\sin x)$ $\frac{dy}{dx} = -2\sin x \cos x \cdot e^{\cos^2 x}$ Using the double angle identity $\sin(2x) = 2\sin x \cos x$: $\frac{dy}{dx} = -\sin(2x) e^{\cos^2 x}$ * **A: $-\sin(2x) e^{\cos^2 x}$** is correct. * **B: $\sin(2x) e^{\cos^2 x}$** is incorrect. This results from a sign error, likely from forgetting the negative sign when differentiating $\cos x$. * **C: $-2\sin(x) e^{\cos^2 x}$** is incorrect. This is missing the $\cos x$ factor from differentiating $\cos^2 x$. * **D: $2\cos(x) e^{\cos^2 x}$** is incorrect. This happens if the derivative of $\cos x$ is forgotten entirely in the inner chain rule step.
6. The function $f$ is given by $f(x) = \frac{e^x}{e^x + 1}$. Find $f'(x)$.
   Answer: $\frac{e^x}{(e^x+1)^2}$ Use the quotient rule, $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$, with $u = e^x$ and $v = e^x + 1$. First, find the derivatives of $u$ and $v$: $u' = e^x$ $v' = e^x$ Apply the quotient rule: $f'(x) = \frac{(e^x)(e^x+1) - (e^x)(e^x)}{(e^x+1)^2}$ Simplify the numerator: $f'(x) = \frac{e^{2x} + e^x - e^{2x}}{(e^x+1)^2}$ $f'(x) = \frac{e^x}{(e^x+1)^2}$ * **A: $\frac{-e^x}{(e^x+1)^2}$** is incorrect. This results from reversing the terms in the numerator of the quotient rule to $uv' - u'v$. * B: 1 is incorrect. This might arise from incorrectly differentiating the numerator and denominator separately and then dividing: $\frac{e^x}{e^x} = 1$. * C: 0 is incorrect. This might arise from an algebraic error in the numerator, such as $e^x(e^x+1) - e^x(e^x+1) = 0$. * **D: $\frac{e^x}{(e^x+1)^2}$** is correct.
7. Find the gradient of the tangent to the curve $y = (x^2 + 1)\sin(x)$ at the point where $x = \pi$.
   Answer: $-(\pi^2+1)$ First, find the derivative $\frac{dy}{dx}$ using the product rule, $(uv)' = u'v + uv'$, with $u = x^2+1$ and $v = \sin(x)$. $u' = 2x$ $v' = \cos(x)$ $\frac{dy}{dx} = (2x)(\sin(x)) + (x^2+1)(\cos(x))$ Now, evaluate the derivative at $x = \pi$. We use the values $\sin(\pi) = 0$ and $\cos(\pi) = -1$. Gradient at $x=\pi$ is: $\frac{dy}{dx}\Big|_{x=\pi} = (2\pi)(\sin(\pi)) + (\pi^2+1)(\cos(\pi))$ $= (2\pi)(0) + (\pi^2+1)(-1)$ $= -(\pi^2+1)$ * **A: $-(\pi^2+1)$** is correct. * **B: $-2\pi$** is incorrect. This comes from an incorrect application of the product rule, for example $u'v' = (2x)(\cos x)$, which evaluates to $2\pi(-1) = -2\pi$ at $x=\pi$. * C: 0 is incorrect. This is the value of the function $y$ at $x=\pi$, not its gradient. * **D: $\pi^2+1$** is incorrect. This results from a sign error, likely by evaluating $\cos(\pi)$ as $+1$ instead of $-1$.
8. The derivative of $f(x) = (ax+b)\cos(x)$ is $f'(x) = 2\cos(x) - (2x+1)\sin(x)$. Find the values of $a$ and $b$.
   Answer: a = 2, b = 1 First, find the derivative of $f(x) = (ax+b)\cos(x)$ using the product rule. Let $u = ax+b$ and $v = \cos(x)$. $u' = a$ $v' = -\sin(x)$ $f'(x) = u'v + uv' = (a)(\cos(x)) + (ax+b)(-\sin(x))$ $f'(x) = a\cos(x) - (ax+b)\sin(x)$ We are given that $f'(x) = 2\cos(x) - (2x+1)\sin(x)$. We can equate the two expressions for $f'(x)$: $a\cos(x) - (ax+b)\sin(x) = 2\cos(x) - (2x+1)\sin(x)$ By comparing the coefficients of the $\cos(x)$ terms, we get: $a = 2$ By comparing the coefficients of the $\sin(x)$ terms (the expressions inside the parentheses), we get: $ax+b = 2x+1$ Substitute $a=2$ into this equation: $2x+b = 2x+1$ This implies that $b=1$. So, $a=2$ and $b=1$. * A: a = -2, b = -1 is incorrect, resulting from sign errors. * B: a = 2, b = -1 is incorrect, resulting from a sign error for b. * C: a = 1, b = 2 is incorrect, mixing up the values of a and b. * D: a = 2, b = 1 is correct.
9. Find the derivative of $f(x) = \frac{\sin x}{1 + \cos x}$.
   Answer: $\frac{1}{1 + \cos x}$ We use the quotient rule with $u = \sin x$ and $v = 1 + \cos x$. $u' = \cos x$ $v' = -\sin x$ $f'(x) = \frac{u'v - uv'}{v^2} = \frac{(\cos x)(1 + \cos x) - (\sin x)(-\sin x)}{(1 + \cos x)^2}$ Simplify the numerator: $f'(x) = \frac{\cos x + \cos^2 x + \sin^2 x}{(1 + \cos x)^2}$ Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$: $f'(x) = \frac{\cos x + 1}{(1 + \cos x)^2}$ Cancel one factor of $(1 + \cos x)$ from the numerator and denominator: $f'(x) = \frac{1}{1 + \cos x}$ * **A: $-\cot x$** is incorrect. This results from the common mistake of differentiating the numerator and denominator separately: $\frac{(\sin x)'}{(1+\cos x)'} = \frac{\cos x}{-\sin x}$. * **B: $\frac{1}{1 + \cos x}$** is correct. * **C: $\frac{-1}{1 + \cos x}$** is incorrect. This could result from a sign error during the simplification of the numerator. * **D: $\frac{\cos x + \cos(2x)}{(1+\cos x)^2}$** is incorrect. This comes from a sign error in the quotient rule formula (using $u'v + uv'$) and then using the double angle identity $\cos(2x) = \cos^2 x - \sin^2 x$.
10. Find the derivative of $f(x) = \cos(\frac{1}{x})$.
    Answer: $\frac{1}{x^2}\sin(\frac{1}{x})$ We must use the chain rule. Let the outer function be $g(u) = \cos(u)$ and the inner function be $u(x) = \frac{1}{x} = x^{-1}$. The derivatives are: $g'(u) = -\sin(u)$ $u'(x) = -1 \cdot x^{-2} = -\frac{1}{x^2}$ By the chain rule, $f'(x) = g'(u(x)) \cdot u'(x)$. $f'(x) = -\sin(\frac{1}{x}) \cdot (-\frac{1}{x^2})$ The two negative signs cancel out: $f'(x) = \frac{1}{x^2}\sin(\frac{1}{x})$ * **A: $-\sin(\frac{1}{x})$** is incorrect. This happens if the derivative of the inner function $\frac{1}{x}$ is forgotten, or assumed to be 1. * **B: $\frac{1}{x^2}\sin(\frac{1}{x})$** is correct. * **C: $-\frac{1}{x^2}\sin(\frac{1}{x})$** is incorrect. This results from making only one sign error, either in differentiating $\cos(u)$ or in differentiating $\frac{1}{x}$, but not both. * **D: $\sin(\frac{1}{x})$** is incorrect. This results from a sign error in the derivative of $\cos(u)$ and also forgetting to differentiate the inner function.
11. Let $f(x) = \sin(\ln x)$ for $x > 0$. Find the value of $f'(e^{\pi/2})$.
    Answer: 0 First, find the derivative $f'(x)$ using the chain rule. Let $u = \ln x$. Then $f(x) = \sin(u)$. $f'(x) = \cos(u) \cdot u' = \cos(\ln x) \cdot \frac{1}{x} = \frac{\cos(\ln x)}{x}$ Now, evaluate this derivative at $x = e^{\pi/2}$. $f'(e^{\pi/2}) = \frac{\cos(\ln(e^{\pi/2}))}{e^{\pi/2}}$ Since $\ln(e^{\pi/2}) = \pi/2$: $f'(e^{\pi/2}) = \frac{\cos(\pi/2)}{e^{\pi/2}}$ We know that $\cos(\pi/2) = 0$, so: $f'(e^{\pi/2}) = \frac{0}{e^{\pi/2}} = 0$ * **A: $-e^{-\pi/2}$** is incorrect. This could arise from miscalculation, perhaps at a different point like $x = e^{\pi}$. * B: 0 is correct. * C: 1 is incorrect. This is the value of the function $f(e^{\pi/2}) = \sin(\ln(e^{\pi/2})) = \sin(\pi/2) = 1$, not its derivative. * **D: $e^{-\pi/2}$** is incorrect. This is the value of $\frac{1}{x}$ at $x = e^{\pi/2}$ if the $\cos(\ln x)$ term in the numerator was mistakenly evaluated as 1.
12. The function $f(x) = \frac{x}{\ln x}$ has a local minimum for $x > 1$. Find the x-coordinate of this minimum.
    Answer: e To find the coordinates of a local minimum, we must first find the derivative of the function and then find the critical points by setting the derivative to zero. Using the quotient rule with $u=x$ and $v=\ln x$: $u'=1$, $v'=1/x$ $f'(x) = \frac{(1)(\ln x) - (x)(\frac{1}{x})}{(\ln x)^2} = \frac{\ln x - 1}{(\ln x)^2}$ Set the derivative to zero to find stationary points: $f'(x) = 0 \implies \frac{\ln x - 1}{(\ln x)^2} = 0$ This requires the numerator to be zero: $\ln x - 1 = 0 \implies \ln x = 1 \implies x = e^1 = e$ To confirm it's a minimum, one could use the second derivative test or analyze the sign of $f'(x)$. For $1 < x < e$, $\ln x < 1$, so $f'(x) < 0$. For $x > e$, $\ln x > 1$, so $f'(x) > 0$. The function is decreasing then increasing, so $x=e$ is a local minimum. * A: 1 is incorrect. The function has a vertical asymptote at $x=1$ because $\ln(1)=0$. A stationary point cannot occur where the function is undefined. * **B: $e^{-1}$** is incorrect. This may arise from an algebraic error like $\ln x = -1$. * C: e is correct. * **D: $e^2$** is incorrect. This may arise from an error like $\ln x = 2$.
13. Let $y = \ln(\sin(e^x))$. Which of the following is $\frac{dy}{dx}$?
    Answer: $e^x \cot(e^x)$ This is a nested chain rule problem. Let's differentiate from the outermost function inwards. 1. The derivative of $\ln(u)$ is $\frac{1}{u}$. Here $u = \sin(e^x)$. 2. The derivative of $\sin(v)$ is $\cos(v)$. Here $v = e^x$. 3. The derivative of $e^x$ is $e^x$. Applying the chain rule, we multiply these derivatives together: $\frac{dy}{dx} = \frac{1}{\sin(e^x)} \cdot \cos(e^x) \cdot e^x$ Simplifying this expression: $\frac{dy}{dx} = e^x \frac{\cos(e^x)}{\sin(e^x)} = e^x \cot(e^x)$ * **A: $\cot(e^x)$** is incorrect. This happens if the last step of the chain rule (differentiating $e^x$) is missed. * **B: $e^x \tan(e^x)$** is incorrect. This uses the wrong trigonometric identity, perhaps from misremembering the derivative of $\ln(\sin u)$. * **C: $\frac{e^x}{\sin(e^x)}$** is incorrect. This happens if the middle step of the chain rule (differentiating $\sin(v)$) is missed. * **D: $e^x \cot(e^x)$** is correct.
14. Find the derivative of $g(x) = \cos^3(2x)$.
    Answer: $-6\sin(2x)\cos^2(2x)$ The function is $g(x) = (\cos(2x))^3$. This requires applying the chain rule twice. Let the outer function be $u^3$, the middle function be $\cos(v)$, and the inner function be $v = 2x$. The derivative is found by differentiating layer by layer from the outside in: 1. Derivative of the power $u^3$ is $3u^2$: $3(\cos(2x))^2$ 2. Derivative of the cosine function $\cos(2x)$ is $-\sin(2x)$ 3. Derivative of the inner function $2x$ is $2$ Multiply these together: $g'(x) = 3(\cos(2x))^2 \cdot (-\sin(2x)) \cdot 2$ $g'(x) = -6\sin(2x)\cos^2(2x)$ * **A: $-3\sin(2x)\cos^2(2x)$** is incorrect. This happens when the derivative of the innermost function $2x$ (which is 2) is forgotten. * **B: $6\sin(2x)\cos^2(2x)$** is incorrect. This is a sign error, forgetting the negative sign from the derivative of $\cos(u)$. * **C: $-6\sin(2x)\cos^2(2x)$** is correct. * **D: $-6\sin^3(2x)$** is incorrect. This confuses which function is which in the chain rule application.
15. Let $h(x) = x \cos(2x^2)$. Find $h'(x)$.
    Answer: $\cos(2x^2) - 4x^2\sin(2x^2)$ We use the product rule, $(uv)' = u'v + uv'$, with $u = x$ and $v = \cos(2x^2)$. $u' = 1$ To find $v'$, we use the chain rule. The derivative of $\cos(w)$ is $-\sin(w) \cdot w'$. Here, $w = 2x^2$, so $w' = 4x$. $v' = -\sin(2x^2) \cdot 4x = -4x\sin(2x^2)$ Now, apply the product rule: $h'(x) = (1)(\cos(2x^2)) + (x)(-4x\sin(2x^2))$ $h'(x) = \cos(2x^2) - 4x^2\sin(2x^2)$ * **A: $\cos(2x^2) - 4x^2\sin(2x^2)$** is correct. * **B: $\cos(2x^2) + 4x^2\sin(2x^2)$** is incorrect. This is a sign error from the derivative of $\cos(w)$, treating it as $\sin(w)$. * **C: $\cos(2x^2) - 2x^2\sin(2x^2)$** is incorrect. This results from an error in the chain rule for $v$, differentiating $2x^2$ as $2x$ instead of $4x$. * **D: $-4x \sin(2x^2)$** is incorrect. This is only the derivative of the second part, $\cos(2x^2)$, and ignores the product rule.
16. Given $y = \frac{\ln x}{x^3}$ for $x > 0$, find $\frac{dy}{dx}$.
    Answer: $\frac{1 - 3\ln x}{x^4}$ We use the quotient rule, $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$, with $u = \ln x$ and $v = x^3$. First, find the derivatives of $u$ and $v$: $u' = \frac{1}{x}$ $v' = 3x^2$ Apply the quotient rule: $\frac{dy}{dx} = \frac{(\frac{1}{x})(x^3) - (\ln x)(3x^2)}{(x^3)^2}$ Now, simplify the expression: $\frac{dy}{dx} = \frac{x^2 - 3x^2 \ln x}{x^6}$ Factor out $x^2$ from the numerator: $\frac{dy}{dx} = \frac{x^2(1 - 3\ln x)}{x^6}$ Cancel the $x^2$ term: $\frac{dy}{dx} = \frac{1 - 3\ln x}{x^4}$ * **A: $\frac{1 - 3\ln x}{x^4}$** is correct. * **B: $\frac{3\ln x - 1}{x^4}$** is incorrect. This results from reversing the terms in the numerator of the quotient rule, calculating $\frac{uv' - u'v}{v^2}$. * **C: $\frac{1 - 3x^2 \ln x}{x^6}$** is incorrect. This is an error in simplifying the term $\frac{1}{x} \cdot x^3$ in the numerator. * **D: $\frac{1 - 3\ln x}{x^3}$** is incorrect. This results from forgetting to square the denominator $v$ in the quotient rule.
17. Let $f(x) = \sqrt{\sin(x) + \cos(x)}$. Find the value of $f'(\frac{\pi}{4})$.
    Answer: 0 First, write the function using a power: $f(x) = (\sin x + \cos x)^{1/2}$. Now, use the chain rule. $f'(x) = \frac{1}{2}(\sin x + \cos x)^{-1/2} \cdot (\cos x - \sin x)$ $f'(x) = \frac{\cos x - \sin x}{2\sqrt{\sin x + \cos x}}$ Now, substitute $x = \frac{\pi}{4}$. At this value, $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$. $f'(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{2\sqrt{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}}$ $f'(\frac{\pi}{4}) = \frac{0}{2\sqrt{\sqrt{2}}} = 0$ * A: 0 is correct. * **B: $\frac{1}{2}$** is incorrect. This might be a guess or result from a calculation error. * **C: $\frac{\sqrt{2}}{2}$** is incorrect. This is the value of $\cos(\frac{\pi}{4})$ and $\sin(\frac{\pi}{4})$. * **D: $2^{1/4}$** is incorrect. This is the value of the original function $f(\frac{\pi}{4}) = \sqrt{\sqrt{2}} = (2^{1/2})^{1/2} = 2^{1/4}$, not its derivative.
18. If $f(x) = \ln(\sqrt{x^2+4})$, what is $f'(x)$?
    Answer: $\frac{x}{x^2+4}$ A helpful first step is to use the properties of logarithms to simplify the function: $f(x) = \ln((x^2+4)^{1/2}) = \frac{1}{2}\ln(x^2+4)$ Now, differentiate using the chain rule. The derivative of $\ln(u)$ is $\frac{u'}{u}$. Here, $u = x^2+4$, so $u' = 2x$. $f'(x) = \frac{1}{2} \cdot \frac{2x}{x^2+4} = \frac{x}{x^2+4}$ Alternatively, without simplifying first, we can use the chain rule on $f(x) = \ln(\sqrt{x^2+4})$. $f'(x) = \frac{1}{\sqrt{x^2+4}} \cdot \frac{d}{dx}(\sqrt{x^2+4})$ $= \frac{1}{\sqrt{x^2+4}} \cdot \frac{1}{2\sqrt{x^2+4}} \cdot (2x)$ $= \frac{2x}{2(x^2+4)} = \frac{x}{x^2+4}$ * **A: $\frac{1}{2(x^2+4)}$** is incorrect. This results from forgetting the $2x$ from the chain rule when differentiating $x^2+4$. * **B: $\frac{x}{x^2+4}$** is correct. * **C: $\frac{x}{\sqrt{x^2+4}}$** is incorrect. This might result from errors in applying the chain rule for the square root and logarithm functions, possibly missing a factor. * **D: $\frac{2x}{\sqrt{x^2+4}}$** is incorrect. This comes from finding the derivative of the inside function, $\sqrt{x^2+4}$, but failing to apply the derivative of the outer $\ln(u)$ function.
19. Let $h(x) = \frac{f(x)}{e^x}$. Given that $f(0)=2$ and $f'(0)=3$, find $h'(0)$.
    Answer: 1 We can find $h'(x)$ using the quotient rule, with $u=f(x)$ and $v=e^x$. $u' = f'(x)$ and $v' = e^x$. $h'(x) = \frac{u'v - uv'}{v^2} = \frac{f'(x)e^x - f(x)e^x}{(e^x)^2}$ Simplifying by cancelling an $e^x$ factor: $h'(x) = \frac{f'(x) - f(x)}{e^x}$ Now, substitute $x=0$ and the given values $f(0)=2$ and $f'(0)=3$: $h'(0) = \frac{f'(0) - f(0)}{e^0} = \frac{3-2}{1} = 1$ Alternatively, one could write $h(x) = f(x)e^{-x}$ and use the product rule: $h'(x) = f'(x)e^{-x} + f(x)(-e^{-x})$ $h'(0) = f'(0)e^0 - f(0)e^0 = 3(1) - 2(1) = 1$ * A: 1 is correct. * B: 2 is incorrect. This is the value of $h(0) = f(0)/e^0 = 2/1$, not its derivative. * C: 3 is incorrect. This might result from an incorrect quotient rule application, for example $h'(x) = f'(x)/e^x$, which gives $h'(0)=f'(0)/e^0 = 3/1 = 3$. * D: 5 is incorrect. This might result from a sign error in the product or quotient rule, leading to $f'(0)+f(0) = 3+2=5$.
20. Find the equation of the normal to the curve $y = \frac{2x}{x^2+1}$ at the origin $(0,0)$.
    Answer: $y = -\frac{1}{2}x$ First, find the gradient of the tangent at $(0,0)$ by finding the derivative $\frac{dy}{dx}$ and evaluating it at $x=0$. Using the quotient rule with $u=2x$ and $v=x^2+1$: $u'=2$, $v'=2x$ $\frac{dy}{dx} = \frac{2(x^2+1) - (2x)(2x)}{(x^2+1)^2} = \frac{2x^2+2-4x^2}{(x^2+1)^2} = \frac{2-2x^2}{(x^2+1)^2}$ At the origin $x=0$, the gradient of the tangent $m_T$ is: $m_T = \frac{2 - 2(0)^2}{(0^2+1)^2} = \frac{2}{1} = 2$ The normal line is perpendicular to the tangent line. Its gradient $m_N$ is the negative reciprocal of the tangent's gradient. $m_N = -\frac{1}{m_T} = -\frac{1}{2}$ The normal line passes through the origin $(0,0)$. The equation of a line through the origin is $y = mx$. Therefore, the equation of the normal is $y = -\frac{1}{2}x$. * **A: $y=2x$** is incorrect. This is the equation of the tangent line, not the normal line. * **B: $y=-2x$** is incorrect. This uses the negative of the tangent gradient, not the negative reciprocal. * **C: $y=\frac{1}{2}x$** is incorrect. This uses the reciprocal of the tangent gradient but omits the negative sign. * **D: $y = -\frac{1}{2}x$** is correct.

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