1.
Question 1. If \(f(x)=\cos(x)\tan(x)\), then \(f'(x)=\)
Step 1: Use the identity \(\tan(x)=\frac{\sin(x)}{\cos(x)}\). Step 2: Rewrite the function: \(f(x)=\cos(x)\cdot\frac{\sin(x)}{\cos(x)}=\sin(x)\). Step 3: Differentiate: \(f'(x)=\cos(x)\). Answer: C. \(\cos(x)\)
2.
Question 3. If \(y=\tan(\sin(x^2))\), then \(y'=\cos(x^2)\sec^2(\sin(x^2))\).
Step 1: Let \(u=\sin(x^2)\), so \(y=\tan(u)\). Step 2: Differentiate the outside function: \(\frac{d}{dx}\tan(u)=\sec^2(u)u'\). Step 3: Differentiate the inside function: \(u'=\cos(x^2)\cdot2x\). Step 4: Therefore, \(y'=2x\cos(x^2)\sec^2(\sin(x^2))\). Step 5: The given derivative is missing the factor \(2x\), so it is false. Answer: B. False
3.
Question 2. If \(f(x)=\cos(3x)\), then \(f'(x)=\)
Step 1: Identify the inside function: \(u=3x\). Step 2: Use the chain rule: \(\frac{d}{dx}\cos(u)=-\sin(u)\cdot u'\). Step 3: Since \(u'=3\), we get \(f'(x)=-\sin(3x)\cdot3\). Step 4: Simplify: \(f'(x)=-3\sin(3x)\). Answer: C. \(-3\sin(3x)\)
4.
Question 4. Find the equation of the line tangent to the graph of \(y=\sin(x)\) at \(x=0\).
Step 1: Differentiate \(y=\sin(x)\): \(y'=\cos(x)\). Step 2: Find the slope at \(x=0\): \(m=\cos(0)=1\). Step 3: Find the point on the curve: \(y=\sin(0)=0\), so the point is \((0,0)\). Step 4: Use point-slope form: \(y-0=1(x-0)\). Step 5: Simplify: \(y=x\). Answer: C. \(y=x\)
5.
Question 6. If \(y=\cos^{-1}(\sqrt{x})\), then \(\frac{dy}{dx}=\frac{1}{2\sqrt{x}\sqrt{1-x}}\).
Step 1: Use the derivative rule \(\frac{d}{dx}\cos^{-1}(u)=-\frac{u'}{\sqrt{1-u^2}}\). Step 2: Let \(u=\sqrt{x}=x^{1/2}\), so \(u'=\frac{1}{2\sqrt{x}}\). Step 3: Substitute into the formula: \(y'=-\frac{\frac{1}{2\sqrt{x}}}{\sqrt{1-(\sqrt{x})^2}}\). Step 4: Simplify: \(y'=-\frac{1}{2\sqrt{x}\sqrt{1-x}}\). Step 5: The given derivative is positive, but the correct derivative is negative, so the statement is false. Answer: A. False
6.
Question 5. What is \(\frac{d}{dx}6^x\)?
Step 1: Use the exponential derivative rule \(\frac{d}{dx}a^x=a^x\ln(a)\). Step 2: Here \(a=6\). Step 3: Therefore, \(\frac{d}{dx}6^x=6^x\ln(6)\). Answer: D. \(6^x\ln(6)\)
7.
Question 7. If \(f(x)=\ln(6x)\), then find \(f'(x)\).
Step 1: Use the chain rule for \(\ln(u)\): \(\frac{d}{dx}\ln(u)=\frac{u'}{u}\). Step 2: Let \(u=6x\), so \(u'=6\). Step 3: Substitute: \(f'(x)=\frac{6}{6x}\). Step 4: Simplify: \(f'(x)=\frac{1}{x}\). Answer: A. \(\frac{1}{x}\)
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Question 8. If \(\cos(xy)=0.5\), then \(y'=-\frac{y}{x}\).
Step 1: Since \(\cos(xy)=0.5\), take inverse cosine of both sides: \(xy=\cos^{-1}(0.5)\). Step 2: The right side is a constant. Step 3: Differentiate both sides with respect to \(x\): \(\frac{d}{dx}(xy)=0\). Step 4: Use the product rule: \(xy'+y=0\). Step 5: Solve for \(y'\): \(xy'=-y\), so \(y'=-\frac{y}{x}\). Step 6: The statement is true. Answer: A. True
9.
Question 9. Find \(\frac{dy}{dx}\), if \(y^4=4x\).
Step 1: Differentiate both sides of \(y^4=4x\) with respect to \(x\). Step 2: Use implicit differentiation on the left side: \(\frac{d}{dx}(y^4)=4y^3\frac{dy}{dx}\). Step 3: Differentiate the right side: \(\frac{d}{dx}(4x)=4\). Step 4: Set them equal: \(4y^3\frac{dy}{dx}=4\). Step 5: Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx}=\frac{4}{4y^3}=\frac{1}{y^3}\). Answer: C. \(\frac{1}{y^3}\)
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