EBUS CAL12 Chapter 3 Differentiation Test-2

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: \(y=11x-14\)

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: \(56x^6\sec^{-1}x+8x^7\frac{1}{|x|\sqrt{x^2-1}}\)

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

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

Step 1: Use the quotient rule with numerator \(u=9x\) and denominator \(v=x+5\). Step 2: Differentiate: \(u'=9\) and \(v'=1\). Step 3: Apply the quotient rule: \(f'(x)=\frac{u'v-uv'}{v^2}\). Step 4: Substitute: \(f'(x)=\frac{9(x+5)-9x(1)}{(x+5)^2}\). Step 5: Simplify the numerator: \(9x+45-9x=45\), so \(f'(x)=\frac{45}{(x+5)^2}\). Step 6: Substitute \(x=5\): \(f'(5)=\frac{45}{(10)^2}=\frac{45}{100}=0.45\). Answer: 0.45

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

Step 1: Let \(u=\frac{4x+2}{2-4x}\). Then \(f(x)=u^2\). Step 2: Use the chain rule: \(f'(x)=2u\cdot u'\). Step 3: Find \(u'\) using the quotient rule. Here \(a=4x+2\), \(b=2-4x\), \(a'=4\), and \(b'=-4\). Step 4: \(u'=\frac{a'b-ab'}{b^2}=\frac{4(2-4x)-(4x+2)(-4)}{(2-4x)^2}\). Step 5: Simplify the numerator: \(8-16x+16x+8=16\), so \(u'=\frac{16}{(2-4x)^2}\). Step 6: Therefore, \(f'(x)=2\left(\frac{4x+2}{2-4x}\right)\left(\frac{16}{(2-4x)^2}\right)=\frac{32(4x+2)}{(2-4x)^3}\). Step 7: Substitute \(x=1\): \(f'(1)=\frac{32(6)}{(-2)^3}=\frac{192}{-8}=-24\). Answer: -24

Step 1: Differentiate \(3^x\): \(\frac{d}{dx}(3^x)=3^x\ln 3\). Step 2: Differentiate \(81\log_4 x\). Since \(\frac{d}{dx}\log_a x=\frac{1}{x\ln a}\), we get \(\frac{d}{dx}(81\log_4 x)=\frac{81}{x\ln 4}\). Step 3: Therefore, \(f'(x)=3^x\ln 3+\frac{81}{x\ln 4}\). Step 4: Substitute \(x=1\): \(f'(1)=3\ln 3+\frac{81}{\ln 4}\). Step 5: Evaluate: \(f'(1)\approx 61.72\). Answer: 61.72

Step 1: Differentiate \(2x^2+y^4=19\) implicitly with respect to \(x\): \(4x+4y^3\frac{dy}{dx}=0\). Step 2: Solve for \(\frac{dy}{dx}\): \(4y^3\frac{dy}{dx}=-4x\), so \(\frac{dy}{dx}=-\frac{x}{y^3}\). Step 3: Differentiate \(\frac{dy}{dx}=-xy^{-3}\) again using the product rule: \(\frac{d^2y}{dx^2}=-\left(y^{-3}+x(-3)y^{-4}\frac{dy}{dx}\right)\). Step 4: Rewrite this as \(\frac{d^2y}{dx^2}=-\frac{1}{y^3}+\frac{3x}{y^4}\frac{dy}{dx}\). Step 5: Substitute \(\frac{dy}{dx}=-\frac{x}{y^3}\): \(\frac{d^2y}{dx^2}=-\frac{1}{y^3}-\frac{3x^2}{y^7}\). Step 6: Substitute \(x=3\) and \(y=1\): \(\frac{d^2y}{dx^2}=-1-27=-28\). Step 7: To two decimal places, the value is \(-28.00\). Answer: -28.00

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: \(\frac{2x^{\ln x}\ln x}{x}\)

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 \(f'(8)\).

Step 1: Start with the definition of the derivative: \(f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\). Step 2: Substitute \(f(x)=\sqrt{x+6}\): \(f'(x)=\lim_{h\to0}\frac{\sqrt{x+h+6}-\sqrt{x+6}}{h}\). Step 3: Multiply by the conjugate: \(\frac{\sqrt{x+h+6}-\sqrt{x+6}}{h}\cdot\frac{\sqrt{x+h+6}+\sqrt{x+6}}{\sqrt{x+h+6}+\sqrt{x+6}}\). Step 4: Simplify the numerator: \((x+h+6)-(x+6)=h\). Step 5: After simplifying, \(f'(x)=\lim_{h\to0}\frac{h}{h(\sqrt{x+h+6}+\sqrt{x+6})}\). Step 6: Cancel \(h\): \(f'(x)=\lim_{h\to0}\frac{1}{\sqrt{x+h+6}+\sqrt{x+6}}\). Step 7: Substitute \(h=0\): \(f'(x)=\frac{1}{\sqrt{x+6}+\sqrt{x+6}}\). Step 8: Simplify: \(f'(x)=\frac{1}{2\sqrt{x+6}}\). Answer: \(\frac{1}{2\sqrt{x+6}}\)