1.
Question 1. Which of the following graphs below best represents the derivative of this graph?
Step 1: The original graph is made of two straight line pieces meeting at \(x=0\).Step 2: For \(x<0\), the graph is increasing with slope \(1\), so the derivative is \(1\).Step 3: For \(x>0\), the graph is decreasing with slope \(-1\), so the derivative is \(-1\).Step 4: At \(x=0\), there is a sharp corner, so the derivative is not defined at that exact point.Step 5: The derivative graph should therefore be a horizontal line at \(y=1\) on the left side and a horizontal line at \(y=-1\) on the right side, with a break at \(x=0\).Answer: Graph B
2.
Question 2. Find the instantaneous slope of \(y=\frac{1}{x}\) at \(x=2\).
Step 1: Rewrite the function as \(y=x^{-1}\).Step 2: Differentiate using the power rule: \(\frac{dy}{dx}=-x^{-2}\).Step 3: Rewrite the derivative as \(\frac{dy}{dx}=-\frac{1}{x^2}\).Step 4: Substitute \(x=2\): \(\left.\frac{dy}{dx}\right|_{x=2}=-\frac{1}{2^2}=-\frac{1}{4}\).Answer: \(-\frac{1}{4}\)
3.
Question 3. Find the equation of the tangent line to \(y=4x\) at \(x=3\).
Step 1: The function is \(y=4x\).Step 2: Differentiate to find the slope of the tangent line: \(\frac{dy}{dx}=4\).Step 3: At \(x=3\), the point on the curve is \(y=4(3)=12\), so the point is \((3,12)\).Step 4: Use point-slope form: \(y-12=4(x-3)\).Step 5: Simplify: \(y-12=4x-12\), so \(y=4x\).Answer: \(y=4x\)
4.
Question 4. If \(h(x)=\frac{x^2}{f(x)}\) and \(f(3)=2\) and \(f'(3)=4\), then \(h'(3)=-12\).
Step 1: Use the quotient rule for \(h(x)=\frac{x^2}{f(x)}\).Step 2: \(\displaystyle h'(x)=\frac{f(x)\cdot 2x-x^2\cdot f'(x)}{[f(x)]^2}\).Step 3: Substitute \(x=3\), \(f(3)=2\), and \(f'(3)=4\).Step 4: \(\displaystyle h'(3)=\frac{2(2)(3)-3^2(4)}{2^2}\).Step 5: Simplify: \(\displaystyle h'(3)=\frac{12-36}{4}=\frac{-24}{4}=-6\).Step 6: Since \(h'(3)=-6\), the statement \(h'(3)=-12\) is false.Answer: False
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Question 5. If \(y=\frac{4x}{x-3}\), then \(\frac{dy}{dx}=\)
Step 1: The function is \(y=\frac{4x}{x-3}\).Step 2: Use the quotient rule: \(\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v u'-u v'}{v^2}\).Step 3: Let \(u=4x\), so \(u'=4\). Let \(v=x-3\), so \(v'=1\).Step 4: Substitute into the quotient rule: \(\displaystyle \frac{dy}{dx}=\frac{(x-3)(4)-(4x)(1)}{(x-3)^2}\).Step 5: Simplify the numerator: \(4x-12-4x=-12\).Step 6: Therefore, \(\displaystyle \frac{dy}{dx}=-\frac{12}{(x-3)^2}\).Answer: \(-\frac{12}{(x-3)^2}\)
6.
Question 6. \(\frac{d}{dx}\left[x^3(x^{99}+1)\right]=3x^2(x^{99}+1)+x^3(99x^{98})\).
Step 1: The expression is a product: \(x^3(x^{99}+1)\).Step 2: Use the product rule: \(\frac{d}{dx}[uv]=u'v+uv'\).Step 3: Let \(u=x^3\), so \(u'=3x^2\). Let \(v=x^{99}+1\), so \(v'=99x^{98}\).Step 4: Substitute into the product rule: \(\frac{d}{dx}\left[x^3(x^{99}+1)\right]=3x^2(x^{99}+1)+x^3(99x^{98})\).Step 5: This matches the given statement, so the statement is true.Answer: True
7.
Question 7. Find \(\frac{dy}{dx}\) if \(y=\sqrt{5}\).
Step 1: The function is \(y=\sqrt{5}\).Step 2: Since \(\sqrt{5}\) is a constant, it does not change as \(x\) changes.Step 3: The derivative of any constant is \(0\).Step 4: Therefore, \(\frac{dy}{dx}=0\).Answer: \(0\)
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