1.
Question 1. Find the equation of the line tangent to \(y=2x^2-x+4\) at \(x=3\).
Step 1: Differentiate the function \(y=2x^2-x+4\) to find the slope function. We get \(y'=4x-1\). Step 2: Substitute \(x=3\) into the derivative: \(y'(3)=4(3)-1=11\). The slope of the tangent line is \(11\). Step 3: Find the point on the original curve when \(x=3\): \(y=2(3)^2-3+4=18-3+4=19\). The point is \((3,19)\). Step 4: Use point-slope form: \(y-y_1=m(x-x_1)\). Step 5: Substitute \(m=11\) and \((x_1,y_1)=(3,19)\): \(y-19=11(x-3)\). Step 6: Simplify: \(y-19=11x-33\), so \(y=11x-14\). Answer: \(y=11x-14\)
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Question 2. Find \(f'(x)\), if \(f(x)=8x^7\sec^{-1}x\).
Step 1: Identify this as a product: \(f(x)=8x^7\sec^{-1}x\). Step 2: Let \(u=8x^7\) and \(v=\sec^{-1}x\). Step 3: Differentiate each part: \(u'=56x^6\) and \(v'=\frac{1}{|x|\sqrt{x^2-1}}\). Step 4: Use the product rule \(f'(x)=u'v+uv'\). Step 5: Substitute: \(f'(x)=56x^6\sec^{-1}x+8x^7\frac{1}{|x|\sqrt{x^2-1}}\). Answer: \(56x^6\sec^{-1}x+8x^7\frac{1}{|x|\sqrt{x^2-1}}\)
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Question 3. The graph of a function \(f\) is shown above. Which of the following could be a graph of \(f'\), the derivative of \(f\)?
Step 1: Look at the behavior of the original graph of \(f\). The graph increases first, then decreases, then increases again. Step 2: Therefore, the derivative \(f'\) must be positive first, then negative, then positive again. Step 3: The original graph has a local maximum at a negative \(x\)-value, so \(f'\) should be \(0\) there. Step 4: The original graph has a local minimum at a positive \(x\)-value, so \(f'\) should also be \(0\) there. Step 5: A graph that is positive-negative-positive and has two zeros should be an upward-opening parabola crossing the \(x\)-axis twice. Step 6: Among the choices, Graph D matches this behavior. Answer: Graph D
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Question 4. True or False: If \(y=-6x^5+\frac{3}{x^{1/2}}\), then \(\frac{d^2y}{dx^2}=-120x^3+\frac{9}{4x^{5/2}}\).
Step 1: Rewrite the function using exponents: \(y=-6x^5+3x^{-1/2}\). Step 2: Find the first derivative: \(y'=-30x^4-\frac{3}{2}x^{-3/2}\). Step 3: Differentiate again to find the second derivative: \(y''=-120x^3+\frac{9}{4}x^{-5/2}\). Step 4: Rewrite \(\frac{9}{4}x^{-5/2}\) as \(\frac{9}{4x^{5/2}}\). Step 5: This matches the given statement. Answer: True
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Question 6. True or False: \(\frac{d}{dx}\sin(\ln(-8x^2))=\cos(\ln(-8x^2))\frac{2}{x}\).
Step 1: Use the chain rule. The outside function is \(\sin(u)\), so the derivative is \(\cos(u)u'\). Step 2: Let \(u=\ln(-8x^2)\). Step 3: Differentiate \(u\): \(u'=\frac{1}{-8x^2}\cdot(-16x)=\frac{2}{x}\). Step 4: Substitute back into the chain rule: \(\frac{d}{dx}\sin(\ln(-8x^2))=\cos(\ln(-8x^2))\frac{2}{x}\). Step 5: The statement is correct. Answer: True
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Question 10. If \(y=x^{\ln x}\), then \(y'= ?\)
Step 1: Use logarithmic differentiation because the exponent contains \(x\). Step 2: Let \(y=x^{\ln x}\). Step 3: Take natural log of both sides: \(\ln y=\ln(x^{\ln x})\). Step 4: Use the power rule for logarithms: \(\ln y=(\ln x)(\ln x)=(\ln x)^2\). Step 5: Differentiate both sides: \(\frac{1}{y}y'=2\ln x\cdot\frac{1}{x}\). Step 6: Multiply by \(y\): \(y'=x^{\ln x}\cdot\frac{2\ln x}{x}\). Step 7: Simplify: \(y'=\frac{2x^{\ln x}\ln x}{x}\). Answer: \(\frac{2x^{\ln x}\ln x}{x}\)
11.
Step 1: A secant line is a line through two points on a graph. In this expression, the two points are \((8,f(8))\) and \((8+h,f(8+h))\). Step 2: The slope of the secant line between these two points is \(\frac{f(8+h)-f(8)}{(8+h)-8}\). Step 3: Simplify the denominator: \((8+h)-8=h\), so the slope is \(\frac{f(8+h)-f(8)}{h}\). Step 4: As \(h\to0\), the point \((8+h,f(8+h))\) moves closer and closer to \((8,f(8))\). Step 5: When the two points become infinitely close, the secant line approaches the tangent line at \(x=8\). Step 6: The derivative \(f'(8)\) is the slope of the tangent line at \(x=8\). Therefore, the limit gives the value of \(f'(8)\). Answer: The limit represents the slope of the tangent line at \(x=8\), so it gives \(f'(8)\).
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