1.
Question 1. At \(x=-2\), the function \(y=\frac{5}{x+2}\) is differentiable but not continuous.
Step 1. A function that is differentiable at a point must also be continuous at that point. Step 2. The statement says the function is differentiable at \(x=-2\) but not continuous at \(x=-2\). This is impossible because differentiability implies continuity. Step 3. Also, \(y=\frac{5}{x+2}\) is undefined at \(x=-2\), so it is not continuous there and cannot be differentiable there. Answer: B. False
2.
Question 2. \(\lim_{h\to 0}\frac{\sqrt{2+h}-\sqrt{2}}{h}\) represents the derivative of \(f(x)=\sqrt{x}\) at \(x=a\). What is \(a\)?
Step 1. The derivative definition is \(f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\). Step 2. For \(f(x)=\sqrt{x}\), this becomes \(f'(a)=\lim_{h\to 0}\frac{\sqrt{a+h}-\sqrt{a}}{h}\). Step 3. Compare this with the given expression \(\lim_{h\to 0}\frac{\sqrt{2+h}-\sqrt{2}}{h}\). Step 4. Since \(a+h=2+h\) and \(a=2\), we get \(a=2\). Answer: D. \(2\)
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Question 3. The derivative of \(y=\sqrt{x}\) is:
Step 1. Rewrite the square root using exponents: \(y=\sqrt{x}=x^{1/2}\). Step 2. Apply the power rule: \(\frac{d}{dx}x^n=nx^{n-1}\). Step 3. Then \(y'=\frac{1}{2}x^{-1/2}\). Step 4. Rewrite \(x^{-1/2}\) as \(\frac{1}{\sqrt{x}}\), so \(y'=\frac{1}{2\sqrt{x}}\). Answer: A. \(y'=\frac{1}{2\sqrt{x}}\)
4.
Question 4. If \(y=\frac{8}{\sqrt{x}}\), find \(y'(4)\).
Step 1. Rewrite the function using exponents: \(y=\frac{8}{\sqrt{x}}=8x^{-1/2}\). Step 2. Differentiate using the power rule: \(y'=8\left(-\frac{1}{2}\right)x^{-3/2}\). Step 3. Simplify: \(y'=-4x^{-3/2}=\frac{-4}{x^{3/2}}\). Step 4. Substitute \(x=4\): \(y'(4)=\frac{-4}{4^{3/2}}\). Step 5. Since \(4^{3/2}=(\sqrt{4})^3=2^3=8\), \(y'(4)=\frac{-4}{8}=-\frac{1}{2}\). Answer: D. \(-\frac{1}{2}\)
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Question 5. If \(f(2)=3\), \(f'(2)=-2\), \(g(2)=4\), and \(g'(2)=-3\), find \(h'(2)\) if \(h(x)=f(x)\times g(x)\).
Step 1. Since \(h(x)=f(x)g(x)\), use the product rule: \(h'(x)=f'(x)g(x)+f(x)g'(x)\). Step 2. Substitute \(x=2\): \(h'(2)=f'(2)g(2)+f(2)g'(2)\). Step 3. Substitute the given values: \(h'(2)=(-2)(4)+(3)(-3)\). Step 4. Simplify: \(h'(2)=-8-9=-17\). Answer: D. \(-17\)
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 of \(x^3\) and \(x^{99}+1\). 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. Answer: B. True
7.
Question 7. If \(y=5x^3+10x^2-15\), find \(\frac{d^2y}{dx^2}\).
Step 1. First find the first derivative of \(y=5x^3+10x^2-15\). Step 2. Differentiate each term: \(\frac{dy}{dx}=15x^2+20x\). Step 3. Now differentiate again to find the second derivative. Step 4. \(\frac{d^2y}{dx^2}=30x+20\). Answer: A. \(30x+20\)
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