EBUS CAL12 Chapter 3 Differentiation Test-1

Step 1: Differentiate the function to find the slope of the tangent line: \(y'=-6x-1\). Step 2: Substitute \(x=1\): \(m=-6(1)-1=-7\). Step 3: Find the point on the curve when \(x=1\): \(y=-3(1)^2-1+9=5\), so the point is \((1,5)\). Step 4: Use point-slope form: \(y-5=-7(x-1)\). Step 5: Simplify: \(y-5=-7x+7\), so \(y=-7x+12\). Answer: \(y=-7x+12\)

Step 1: Use the product rule with \(u=9x^3\) and \(v=\sec^{-1}x\). Step 2: Differentiate each part: \(u'=27x^2\) and \(v'=\frac{1}{|x|\sqrt{x^2-1}}\). Step 3: Apply \(f'(x)=u'v+uv'\). Step 4: Substitute the derivatives: \(f'(x)=27x^2\sec^{-1}x+9x^3\frac{1}{|x|\sqrt{x^2-1}}\). Answer: \(27x^2\sec^{-1}x+9x^3\frac{1}{|x|\sqrt{x^2-1}}\)

Step 1: Look at where the original graph is increasing and decreasing. It increases first, then decreases, then increases again. Step 2: Therefore, the derivative must be positive, then negative, then positive. Step 3: The original graph has a local maximum at a negative \(x\)-value and a local minimum at a positive \(x\)-value, so \(f'\) should have two zeros at those points. Step 4: A derivative graph with positive-negative-positive behavior is an upward-opening parabola crossing the \(x\)-axis twice. Step 5: Among the choices, Graph A matches this behavior. Answer: Graph A

Step 1: Rewrite the function using exponents: \(y=2x^8+7x^{-1/2}\). Step 2: Find the first derivative: \(y'=16x^7-\frac{7}{2}x^{-3/2}\). Step 3: Differentiate again: \(y''=112x^6+\frac{21}{4}x^{-5/2}\). Step 4: Rewrite \(\frac{21}{4}x^{-5/2}\) as \(\frac{21}{4x^{5/2}}\). Step 5: The given second derivative is correct. Answer: True

Step 1: Use the product rule with \(u=-5x^2-2x\) and \(v=2x^2+4\). Step 2: Differentiate each part: \(u'=-10x-2\) and \(v'=4x\). Step 3: Apply the product rule: \(f'(x)=u'v+uv'\). Step 4: Substitute: \(f'(x)=(-10x-2)(2x^2+4)+(-5x^2-2x)(4x)\). Step 5: Expand and simplify: \(f'(x)=-40x^3-12x^2-40x-8\). Step 6: Evaluate at \(x=-2\): \(f'(-2)=-40(-2)^3-12(-2)^2-40(-2)-8=320-48+80-8=344\). Answer: 344

Step 1: Use the chain rule. The outside function is \(\sin(u)\), so its derivative is \(\cos(u)u'\). Step 2: Let \(u=\ln(-8x^8)\). Step 3: Differentiate \(u\): \(u'=\frac{1}{-8x^8}(-64x^7)=\frac{8}{x}\). Step 4: Substitute into the chain rule: \(\frac{d}{dx}\sin(\ln(-8x^8))=\cos(\ln(-8x^8))\frac{8}{x}\). Step 5: The statement is correct. Answer: True

Step 1: Let \(u=\frac{2-3x}{3x+2}\), so \(f(x)=u^2\). Step 2: Use the chain rule: \(f'(x)=2u\cdot u'\). Step 3: Find \(u'\) using the quotient rule: \(u'=\frac{(3x+2)(-3)-(2-3x)(3)}{(3x+2)^2}\). Step 4: Simplify the numerator: \(-9x-6-6+9x=-12\), so \(u'=\frac{-12}{(3x+2)^2}\). Step 5: Therefore, \(f'(x)=2\left(\frac{2-3x}{3x+2}\right)\left(\frac{-12}{(3x+2)^2}\right)=\frac{-24(2-3x)}{(3x+2)^3}\). Step 6: Substitute \(x=-1\): \(f'(-1)=\frac{-24(2-3(-1))}{(3(-1)+2)^3}=\frac{-24(5)}{(-1)^3}=120\). Answer: 120

Step 1: Differentiate \(4^x\): \(\frac{d}{dx}(4^x)=4^x\ln 4\). 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, \(f'(x)=4^x\ln 4+\frac{81}{x\ln 4}\). Step 4: Substitute \(x=2\): \(f'(2)=4^2\ln 4+\frac{81}{2\ln 4}\). Step 5: Evaluate: \(f'(2)\approx 51.40\). Answer: 51.40

Step 1: Differentiate \(2x^2+y^2=17\) implicitly: \(4x+2y\frac{dy}{dx}=0\). Step 2: Solve for the first derivative: \(\frac{dy}{dx}=\frac{-4x}{2y}=\frac{-2x}{y}\). Step 3: Differentiate \(\frac{dy}{dx}=\frac{-2x}{y}\) again using the quotient rule: \(\frac{d^2y}{dx^2}=\frac{-2y+2x\frac{dy}{dx}}{y^2}\). Step 4: At \((x,y)=(2,1)\), first find \(\frac{dy}{dx}=\frac{-2(2)}{1}=-4\). Step 5: Substitute into the second derivative: \(\frac{d^2y}{dx^2}=\frac{-2(1)+2(2)(-4)}{1^2}=-2-16=-18\). Step 6: Using the answer key provided with the uploaded file, the listed answer is \(-68.00\), but differentiating \(2x^2+y^2=17\) gives \(-18.00\). Answer: -18.00

Step 1: Use logarithmic differentiation because both the base and exponent involve \(x\). Step 2: Let \(y=(x^2+1)^x\). Step 3: Take natural log of both sides: \(\ln y=x\ln(x^2+1)\). Step 4: Differentiate implicitly: \(\frac{1}{y}y'=\ln(x^2+1)+x\cdot\frac{2x}{x^2+1}\). Step 5: Simplify: \(\frac{1}{y}y'=\ln(x^2+1)+\frac{2x^2}{x^2+1}\). Step 6: Multiply by \(y\): \(y'=(x^2+1)^x\left(\ln(x^2+1)+\frac{2x^2}{x^2+1}\right)\). Answer: \((x^2+1)^x\left(\ln(x^2+1)+\frac{2x^2}{x^2+1}\right)\)

Step 1: A secant line connects two points on the graph, such as \((2,f(2))\) and \((2+h,f(2+h))\). Step 2: The slope of this secant line is \(\frac{f(2+h)-f(2)}{(2+h)-2}=\frac{f(2+h)-f(2)}{h}\). Step 3: As \(h\to0\), the second point \((2+h,f(2+h))\) moves closer and closer to \((2,f(2))\). Step 4: When the two points become infinitely close, the secant line approaches the tangent line at \(x=2\). Step 5: The slope of the tangent line is the derivative at that point. Step 6: Therefore, \(\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=f'(2)\). Answer: The limit gives the slope of the tangent line to \(f(x)\) at \(x=2\), which is \(f'(2)\).

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: Cancel \(h\): \(f'(x)=\lim_{h\to0}\frac{1}{\sqrt{x+h+6}+\sqrt{x+6}}\). Step 6: Substitute \(h=0\): \(f'(x)=\frac{1}{\sqrt{x+6}+\sqrt{x+6}}=\frac{1}{2\sqrt{x+6}}\). Answer: \(f'(x)=\frac{1}{2\sqrt{x+6}}\)