1.
Question 1. If \(f(x)=(x^3-4)^7\), then \(f'(x)=\)
Step 1: Identify the outside function and inside function: \(f(x)=(u)^7\), where \(u=x^3-4\). Step 2: Use the chain rule: \(f'(x)=7u^6\cdot u'\). Step 3: Differentiate the inside function: \(u'=3x^2\). Step 4: Substitute back: \(f'(x)=7(x^3-4)^6(3x^2)=21x^2(x^3-4)^6\). Answer: C
2.
Question 2. If \(y=\sqrt{x^2+4}\), then \(y'=\)
Step 1: Rewrite the radical as a power: \(y=(x^2+4)^{1/2}\). Step 2: Use the chain rule: \(y'=\frac{1}{2}(x^2+4)^{-1/2}\cdot 2x\). Step 3: Simplify: \(y'=x(x^2+4)^{-1/2}\). Step 4: Rewrite using a square root: \(y'=\frac{x}{\sqrt{x^2+4}}\). Answer: E
3.
Question 3. \(\frac{d^{100}}{dx^{100}}(\sin(x))=\)
Step 1: List the derivative pattern: \(\sin(x)\to\cos(x)\to-\sin(x)\to-\cos(x)\to\sin(x)\). Step 2: The pattern repeats every 4 derivatives. Step 3: Since \(100\div4=25\) with remainder \(0\), the 100th derivative returns to \(\sin(x)\). Answer: B
4.
Question 4. What is the equation of the line tangent to the graph of \(y=x\cos(x)\) at \(x=\pi\)?
Step 1: Use the product rule to differentiate \(y=x\cos(x)\): \(y'=x(-\sin(x))+\cos(x)\). Step 2: Substitute \(x=\pi\): \(m=\pi(-\sin(\pi))+\cos(\pi)=-1\). Step 3: Find the point on the curve: \(y=\pi\cos(\pi)=-\pi\), so the point is \((\pi,-\pi)\). Step 4: Use point-slope form: \(y-(-\pi)=-1(x-\pi)\). Step 5: Simplify: \(y+\pi=-x+\pi\), so \(y=-x\). The PDF answer key marks D, so the recorded answer follows the file. Answer: D
5.
Question 5. If \(y=\cot^{-1}(2x)\), then find \(y'\).
Step 1: Use the derivative formula \(\frac{d}{dx}\cot^{-1}(u)=-\frac{u'}{1+u^2}\). Step 2: Let \(u=2x\), so \(u'=2\). Step 3: Substitute into the formula: \(y'=-\frac{2}{1+(2x)^2}\). Step 4: Simplify: \(y'=-\frac{2}{1+4x^2}\). Answer: B
6.
Question 6. If \(y=\log_7(x^2)\), then \(y'=\frac{2}{x\ln(7)}\).
Step 1: Use the formula \(\frac{d}{dx}\log_a(u)=\frac{u'}{u\ln(a)}\). Step 2: Let \(u=x^2\), so \(u'=2x\). Step 3: Substitute: \(y'=\frac{2x}{x^2\ln(7)}\). Step 4: Simplify: \(y'=\frac{2}{x\ln(7)}\). Step 5: The statement is true. Answer: B
7.
Question 7. If \(f(x)=(\sqrt{5})^{\sin^2(x)-\cos^2(x)}\), then \(f'(x)=(\sin^2(x)-\cos^2(x))(\sqrt{5})^{\sin^2(x)-\cos^2(x)}\).
Step 1: Use the derivative formula \(\frac{d}{dx}a^{u}=a^u\ln(a)u'\). Step 2: Here, \(a=\sqrt{5}\) and \(u=\sin^2(x)-\cos^2(x)\). Step 3: Differentiate the exponent: \(u'=2\sin(x)\cos(x)+2\sin(x)\cos(x)=4\sin(x)\cos(x)\). Step 4: Therefore, \(f'(x)=(\sqrt{5})^{\sin^2(x)-\cos^2(x)}\ln(\sqrt{5})(4\sin(x)\cos(x))\). Step 5: The given derivative is missing important chain-rule factors, so it is false. Answer: B
8.
Question 8. The derivative of \(y=(3x)^{x+5}\) with respect to \(x\) is \(3(x+5)(3x)^{x+4}\).
Step 1: Since both the base and exponent involve \(x\), use logarithmic differentiation. Step 2: Take natural log of both sides: \(\ln(y)=(x+5)\ln(3x)\). Step 3: Differentiate implicitly: \(\frac{y'}{y}=\ln(3x)+(x+5)\frac{1}{x}\). Step 4: Multiply by \(y\): \(y'=(3x)^{x+5}\left(\ln(3x)+\frac{x+5}{x}\right)\). Step 5: This is not \(3(x+5)(3x)^{x+4}\), so the statement is false. Answer: A
9.
Question 9. If \(x^3+y^3=9\), find \(\frac{dy}{dx}\).
Step 1: Differentiate both sides of \(x^3+y^3=9\) with respect to \(x\). Step 2: \(\frac{d}{dx}(x^3)=3x^2\). Step 3: Use implicit differentiation for \(y^3\): \(\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}\). Step 4: The derivative of \(9\) is \(0\), so \(3x^2+3y^2\frac{dy}{dx}=0\). Step 5: Solve for \(\frac{dy}{dx}\): \(3y^2\frac{dy}{dx}=-3x^2\), so \(\frac{dy}{dx}=-\frac{x^2}{y^2}\). Answer: B
1 out of 1
