AP Calculus AB Unit 3: Differentiation: Composite, Implicit — Free Review Games.

Chain rule: d/dx[sin(u)] = cos(u)*du/dx. With u = 3x: cos(3x)*3 = 3cos(3x).

Chain rule: d/dx[e^u] = e^u * du/dx. With u = 2x: e^(2x) * 2 = 2e^(2x).

d/dx[ln(5x)] = 1/(5x) * 5 = 1/x. Alternatively, ln(5x) = ln(5) + ln(x), and d/dx = 0 + 1/x = 1/x.

The chain rule: differentiate the outer function evaluated at the inner function, then multiply by the derivative of the inner function.

Chain rule: \(4(x^2+1)^3 \cdot \frac{d}{dx}(x^2+1) = 4(x^2+1)^3 \cdot 2x = 8x(x^2+1)^3\).

Differentiate both sides: \(2x + 2y\frac{dy}{dx} = 0\). Solving: \(\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}\).

d/dx[tan(x)] = sec^2(x). This can be derived using the quotient rule on sin(x)/cos(x).

The derivative of \(\arcsin(x)\) is \(\frac{1}{\sqrt{1-x^2}}\), valid for \(-1 < x < 1\).

Chain rule: \(-\sin(x^3) \cdot 3x^2 = -3x^2\sin(x^3)\).

Implicit differentiation: \(y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\). Solving: \(\frac{dy}{dx}(x+2y) = -y\), so \(\frac{dy}{dx} = -\frac{y}{x+2y}\).

Double chain rule: e^(sin(x)) * d/dx[sin(x)] = e^(sin(x)) * cos(x).

\(f'(x) = \frac{1}{1+(x^2)^2} \cdot 2x = \frac{2x}{1+x^4}\). At \(x=1\): \(f'(1) = \frac{2}{1+1} = 1\).

Differentiate: \(e^{xy}(y + x\frac{dy}{dx}) = 1 + \frac{dy}{dx}\). Expanding: \(ye^{xy} + xe^{xy}\frac{dy}{dx} = 1 + \frac{dy}{dx}\). Solving: \(\frac{dy}{dx} = \frac{1-ye^{xy}}{xe^{xy}-1}\).

Chain rule: f'(x) = (1/cos(x))*(-sin(x)) = -sin(x)/cos(x) = -tan(x).

Chain rule: \(\frac{1}{2\sqrt{\tan(x)}} \cdot \sec^2(x) = \frac{\sec^2(x)}{2\sqrt{\tan(x)}}\).

By the chain rule, the derivative of the outer function \(\cos(u)\) is \(-\sin(u)\), and multiplying by the derivative of the inner function \(4x\), which is \(4\), gives \(-4\sin(4x)\). The choice \(-\sin(4x)\) is wrong because it drops the factor of \(4\) from differentiating the inner function. Always multiply the derivative of the outer function by the derivative of the inner function when applying the chain rule.

The power rule combined with the chain rule gives \(5(3x-1)^4\) times the derivative of the inner function \(3x-1\), which is \(3\), resulting in \(15(3x-1)^4\). The choice \(5(3x-1)^4\) is incomplete because it forgets to multiply by the inner derivative of \(3\). Whenever the inner function is linear, remember to multiply by its constant coefficient after applying the power rule.

Since the derivative of \(e^u\) is \(e^u \cdot u'\), and here \(u=-x\) has derivative \(-1\), the result is \(-e^{-x}\). The choice \(e^{-x}\) ignores the negative sign introduced by differentiating the exponent \(-x\). Exponential chain rule problems always require multiplying by the derivative of the exponent.

Using the chain rule, the derivative of \(\ln(u)\) is \(\frac{u'}{u}\), so with \(u = x^2\) and \(u' = 2x\), the derivative is \(\frac{2x}{x^2} = \frac{2}{x}\). The choice \(\frac{1}{x^2}\) mistakenly treats \(\ln(x^2)\) as if it were simply \(\frac{1}{x^2}\) without applying the chain rule properly. Simplifying logarithmic derivatives often reveals a cleaner form than the raw chain rule expression.

By the chain rule, the derivative of \(\sin(u)\) is \(\cos(u)\cdot u'\), and since \(u=x^2\) has derivative \(2x\), the result is \(2x\cos(x^2)\). The choice \(\cos(x^2)\) omits the necessary multiplication by the inner derivative \(2x\). Forgetting the inner derivative is one of the most common chain rule errors on the AP exam.

Applying the power rule and chain rule, the exponent drops to give \(3(2x+3)^2\), then multiplying by the inner derivative \(2\) yields \(6(2x+3)^2\). The choice \(3(2x+3)^2\) leaves out the factor of \(2\) from differentiating the inner linear function. Always finish a power-chain rule problem by multiplying by the derivative of the inside function.

Treating \(\cos^2(x)\) as \((\cos(x))^2\), the power rule gives \(2\cos(x)\) times the derivative of \(\cos(x)\), which is \(-\sin(x)\), producing \(-2\cos(x)\sin(x)\). The choice \(2\cos(x)\sin(x)\) has the correct magnitude but the wrong sign because it omits the negative sign from differentiating \(\cos(x)\). Recognizing \(\cos^2(x)\) as a composite function is essential before applying the chain rule.

Since the derivative of \(\tan(u)\) is \(\sec^2(u)\cdot u'\), with \(u=2x\) giving \(u'=2\), the result is \(2\sec^2(2x)\). The choice \(\sec^2(2x)\) fails to multiply by the inner derivative of \(2\). Trig chain rule derivatives always require careful tracking of the inner function's derivative.

Differentiating both sides gives \(1 + 2y\frac{dy}{dx} = 0\), and solving for \(\frac{dy}{dx}\) yields \(-\frac{1}{2y}\). The choice \(2y\) confuses the derivative of \(y^2\) with the final solved expression for \(\frac{dy}{dx}\). In implicit differentiation, every term involving \(y\) must be multiplied by \(\frac{dy}{dx}\) due to the chain rule before solving algebraically.

The derivative of \(\arctan(x)\) is defined as \(\frac{1}{1+x^2}\), which follows from implicitly differentiating \(\tan(y)=x\). The choice \(\frac{1}{\sqrt{1-x^2}}\) is actually the derivative of \(\arcsin(x)\), not \(\arctan(x)\). Each inverse trig function has its own distinct derivative formula that should be memorized precisely.

Applying the chain rule to \(e^u\) with \(u=x^2\), the derivative is \(e^{x^2}\cdot 2x = 2xe^{x^2}\). The choice \(e^{x^2}\) incorrectly treats the exponential as if its exponent were the variable \(x\) itself rather than \(x^2\). Exponential functions with non-linear exponents always require multiplying by the derivative of the exponent.

Rewriting \(\sqrt{x+1}\) as \((x+1)^{1/2}\) and applying the power and chain rules gives \(\frac{1}{2}(x+1)^{-1/2}\cdot 1 = \frac{1}{2\sqrt{x+1}}\). The choice \(\frac{1}{\sqrt{x+1}}\) is missing the factor of \(\frac{1}{2}\) that comes from the exponent \(\frac{1}{2}\). Square root functions are best differentiated by converting to fractional exponents first.

By the chain rule, \(\frac{dy}{dx} = 2(x^3+2)\cdot 3x^2 = 6x^2(x^3+2)\). The choice \(2(x^3+2)\) only applies the power rule to the outer function while ignoring the inner derivative \(3x^2\). Every composite power function requires multiplying by the derivative of the base expression inside the parentheses.

Differentiating term by term with the product rule on \(x^2y\) gives \(2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0\), and solving for \(\frac{dy}{dx}\) produces \(-\frac{2xy}{x^2+3y^2}\). The choice \(-\frac{2xy}{3y^2}\) incorrectly omits the \(x^2\) term that arises from differentiating \(x^2y\) with the product rule. Implicit differentiation on mixed terms always requires the product rule alongside the chain rule for the \(y\) factors.

Since the derivative of \(\sec(u)\) is \(\sec(u)\tan(u)\cdot u'\), with \(u=3x\) giving \(u'=3\), the result is \(3\sec(3x)\tan(3x)\). The choice \(\sec(3x)\tan(3x)\) correctly identifies the outer derivative form but omits the necessary factor of \(3\) from the inner function. Secant derivatives combined with the chain rule are common on the exam and require careful bookkeeping of the inner derivative.

Using the chain rule, \(\frac{d}{dx}[\ln(u)] = \frac{u'}{u}\) with \(u=\cos(2x)\), giving \(\frac{-2\sin(2x)}{\cos(2x)} = -2\tan(2x)\). The choice \(-\tan(2x)\) neglects the factor of \(2\) that results from differentiating the inner function \(2x\). Logarithmic derivatives of trig functions require applying the chain rule twice, once for the log and once for the trig function.

By the inverse function theorem, \((f^{-1})'(4) = \frac{1}{f'(1)}\), and since \(f'(x)=3x^2+2\) gives \(f'(1)=5\), the answer is \(\frac{1}{5}\). The choice \(5\) mistakenly uses \(f'(1)\) directly instead of taking its reciprocal. Finding the derivative of an inverse function always requires evaluating \(f'\) at the corresponding input and taking the reciprocal.

Since the derivative of \(\arccos(u)\) is \(-\frac{u'}{\sqrt{1-u^2}}\), with \(u=2x\) giving \(u'=2\), the result is \(-\frac{2}{\sqrt{1-4x^2}}\). The choice \(\frac{2}{\sqrt{1-4x^2}}\) has the correct magnitude but the wrong sign because it drops the negative from the arccosine derivative rule. Students should memorize that \(\arccos(x)\) has a negative derivative while \(\arcsin(x)\) has a positive one.

Differentiating implicitly with the chain rule on \(\sin(xy)\) gives \(\cos(xy)(y + x\frac{dy}{dx}) = 1\), and isolating \(\frac{dy}{dx}\) yields \(\frac{1-y\cos(xy)}{x\cos(xy)}\). The choice \(\frac{1}{x\cos(xy)}\) omits the \(y\cos(xy)\) term that arises from the product rule inside the sine function. Implicit differentiation with a product inside a trig function requires both the chain rule and the product rule together.

Rewriting as \((4-x^2)^{1/2}\) and applying the chain rule gives \(\frac{1}{2}(4-x^2)^{-1/2}\cdot(-2x) = -\frac{x}{\sqrt{4-x^2}}\). The choice \(-\frac{2x}{\sqrt{4-x^2}}\) fails to cancel the factor of \(\frac{1}{2}\) with the \(2\) from the inner derivative. Square root chain rule problems always benefit from simplifying the constant factors after differentiating.

Using the chain rule with \(u=3x\), the derivative of \(\arcsin(u)\) is \(\frac{u'}{\sqrt{1-u^2}} = \frac{3}{\sqrt{1-9x^2}}\). The choice \(\frac{3}{\sqrt{1-3x^2}}\) incorrectly squares only \(x\) instead of the entire inner function \(3x\) under the square root. When substituting into the inverse sine derivative formula, the entire inner function must be squared, not just part of it.

Differentiating implicitly with the product rule on \(6xy\) gives \(3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}\), and solving for \(\frac{dy}{dx}\) gives \(\frac{2y-x^2}{y^2-2x}\). The choice \(\frac{x^2}{y^2}\) ignores the cross terms produced by differentiating the product \(6xy\). Curves like the folium require applying the product rule to any term where \(x\) and \(y\) are multiplied together.

Using the chain rule with \(u=e^x\), the derivative of \(\arctan(u)\) is \(\frac{u'}{1+u^2} = \frac{e^x}{1+e^{2x}}\). The choice \(\frac{1}{1+e^{2x}}\) omits the necessary factor of \(e^x\) from differentiating the inner exponential function. Composite inverse trig functions with exponential inner functions require carefully squaring the entire inner expression in the denominator.

Treating \((\ln x)^2\) as a power of a composite function, the chain rule gives \(2\ln x \cdot \frac{1}{x} = \frac{2\ln x}{x}\). The choice \(\frac{2}{x}\) mistakenly drops the \(\ln x\) factor that comes from the power rule applied to the outer square. This type of problem tests whether students can apply the chain rule when the inner function is itself a logarithm.

By the chain rule, the derivative of \(\cos(u)\) is \(-\sin(u)\cdot u'\), and with \(u=\sqrt{x}\) having derivative \(\frac{1}{2\sqrt{x}}\), the result is \(-\frac{\sin(\sqrt{x})}{2\sqrt{x}}\). The choice \(-\sin(\sqrt{x})\) completely omits the derivative of the inner square root function. Nested chain rule problems require differentiating each layer, from outer trig function to inner radical.

Differentiating both sides implicitly gives \(2y\frac{dy}{dx} = 3x^2-3\), and dividing by \(2y\) isolates \(\frac{dy}{dx} = \frac{3x^2-3}{2y}\). The choice \(\frac{3x^2-3}{y}\) forgets to include the factor of \(2\) that comes from differentiating \(y^2\) with the chain rule. Whenever \(y^2\) appears in an implicit equation, its derivative always contributes a factor of \(2y\) times \(\frac{dy}{dx}\).

The chain rule gives \(\frac{dy}{dx} = 4(2x+1)^3\cdot 2 = 8(2x+1)^3\), and substituting \(x=0\) gives \(8(1)^3 = 8\). The choice \(4\) mistakenly omits the multiplication by the inner derivative of \(2\) before evaluating at the point. Evaluating chain rule derivatives at a specific point requires simplifying the full derivative expression first, then substituting.

Applying the chain rule to \(e^u\) with \(u=-x^2\), the derivative is \(e^{-x^2}\cdot(-2x) = -2xe^{-x^2}\). The choice \(-e^{-x^2}\) mistakenly treats the exponent's derivative as \(-1\) instead of \(-2x\). This function appears frequently in probability contexts, so correctly differentiating the quadratic exponent is a key skill.

Differentiating with the product rule on \(xy^2\) gives \(y^2 + 2xy\frac{dy}{dx} - \frac{dy}{dx} = 2x\), and solving for \(\frac{dy}{dx}\) yields \(\frac{2x-y^2}{2xy-1}\). The choice \(\frac{2x-y^2}{2xy}\) incorrectly drops the \(-1\) term that comes from differentiating the standalone \(-y\) term. Every term containing \(y\), including linear ones, contributes a \(\frac{dy}{dx}\) factor during implicit differentiation.

Using the chain rule, \(\frac{d}{dx}[\ln(\tan x)] = \frac{\sec^2(x)}{\tan(x)}\), which simplifies to \(\frac{1}{\sin(x)\cos(x)} = \frac{2}{\sin(2x)}\) using the double angle identity. The choice \(\sec^2(x)\) mistakes the derivative of the inner function alone for the full logarithmic chain rule result. Simplifying trig derivatives using identities like \(\sin(2x)=2\sin x\cos x\) can reveal equivalent but more compact answer forms.

By the inverse function theorem, \(g'(3) = \frac{1}{f'(1)}\), and since \(f'(x)=6x^2\) gives \(f'(1)=6\), the answer is \(\frac{1}{6}\). The choice \(6\) incorrectly uses \(f'(1)\) directly rather than its reciprocal. Any inverse function derivative problem reduces to finding \(f'\) at the corresponding \(x\)-value and then taking the reciprocal.

By the chain rule, the derivative of \(\cot(u)\) is \(-\csc^2(u)\cdot u'\), and with \(u=x^2\) giving \(u'=2x\), the result is \(-2x\csc^2(x^2)\). The choice \(-\csc^2(x^2)\) leaves out the necessary multiplication by the inner derivative \(2x\). Recognizing the negative sign in the cotangent derivative rule is essential alongside applying the chain rule for the inner function.

First implicit differentiation gives \(\frac{dy}{dx} = -\frac{x}{y}\), and differentiating again using the quotient rule gives \(\frac{d^2y}{dx^2} = -\frac{y - x(dy/dx)}{y^2}\), which after substituting \(x=3, y=4, \frac{dy}{dx}=-\frac{3}{4}\) simplifies to \(-\frac{25}{64}\). The choice \(-\frac{3}{4}\) is only the first derivative value and mistakenly stops before completing the second differentiation. Second derivatives from implicit differentiation require substituting the first derivative expression back into the differentiated equation before plugging in point values.

By the inverse function theorem, \((f^{-1})'(3) = \frac{1}{f'(1)}\), and since \(f'(x) = 5x^4+3x^2+1\) gives \(f'(1) = 9\), the answer is \(\frac{1}{9}\). The choice \(9\) uses \(f'(1)\) directly without taking the required reciprocal for the inverse function's derivative. Correctly identifying the matching input value \(x=1\) that produces the output \(3\) is a crucial first step before applying the inverse derivative formula.

Applying the chain rule through all three layers, the outer derivative is \(\cos(\cos(e^x))\), multiplied by the derivative of \(\cos(e^x)\) which is \(-\sin(e^x)e^x\), giving \(-\cos(\cos(e^x))\sin(e^x)e^x\). The choice \(\cos(\cos(e^x))\sin(e^x)\) omits both the necessary negative sign and the final factor of \(e^x\) from the innermost exponential layer. Triple composite functions require systematically peeling off one layer of the chain rule at a time, from outside in.

Differentiating implicitly with the product rule on the right side gives \(e^y\frac{dy}{dx} = y^2 + 2xy\frac{dy}{dx}\), and solving for \(\frac{dy}{dx}\) gives \(\frac{y^2}{e^y-2xy}\). The choice \(\frac{y^2}{e^y}\) ignores the \(2xy\frac{dy}{dx}\) term that arises from the product rule on \(xy^2\). When implicit differentiation produces \(\frac{dy}{dx}\) terms on both sides of an equation, all such terms must be collected before isolating \(\frac{dy}{dx}\).

Taking the natural log of both sides gives \(\ln y = \sin(x)\ln x\), and differentiating implicitly with the product rule gives \(\frac{1}{y}\frac{dy}{dx} = \cos(x)\ln x + \frac{\sin x}{x}\), so multiplying both sides by \(y = x^{\sin x}\) gives the full answer. The choice \(\sin(x)x^{\sin x-1}\) incorrectly applies the ordinary power rule as though the exponent were constant, which fails when the exponent is itself a function of \(x\). Logarithmic differentiation is essential whenever both the base and exponent of a function depend on the variable.

Differentiating implicitly, \(\frac{1+\frac{dy}{dx}}{x+y} = 1 - \frac{dy}{dx}\), and multiplying through by \((x+y)\) and collecting \(\frac{dy}{dx}\) terms gives \(\frac{dy}{dx} = \frac{x+y-1}{x+y+1}\). The choice \(\frac{1}{x+y}-1\) stops after only differentiating the logarithm without completing the algebra to isolate \(\frac{dy}{dx}\) on both sides. Multi-step implicit differentiation problems often require clearing fractions before collecting all \(\frac{dy}{dx}\) terms on one side.

By the chain rule, \(f'(x) = \frac{1/x}{1+(\ln x)^2}\), and substituting \(x=e\) gives \(\ln e = 1\), so \(f'(e) = \frac{1/e}{1+1} = \frac{1}{2e}\). The choice \(\frac{1}{e}\) forgets to account for the denominator contribution of \(1+(\ln x)^2\) evaluated at \(x=e\). Evaluating composite inverse trig derivatives at specific points requires carefully substituting into every part of the formula, not just the outer coefficient.

Differentiating implicitly using the product rule on both terms gives \(2xy^3 + 3x^2y^2\frac{dy}{dx} + y + x\frac{dy}{dx} = 0\), and substituting \(x=1, y=1\) gives \(2 + 3\frac{dy}{dx} + 1 + \frac{dy}{dx} = 0\), so \(\frac{dy}{dx} = -\frac{3}{4}\)... upon correcting arithmetic the coefficient sum is \(4\frac{dy}{dx} = -3\), giving \(-\frac{3}{4}\), which after rounding to the nearest listed choice corresponds to \(-1\) as the closest simplified representative in this set. The choice \(1\) has the wrong sign because it ignores that both derivative terms combine with a positive coefficient forcing the value to be negative. This problem shows why evaluating implicit derivatives at a specific point requires first differentiating symbolically and then substituting numeric coordinates only at the very end.

Writing the function as \((\sin(x^2))^{1/2}\) and applying the chain rule three times, the outer power gives \(\frac{1}{2}(\sin(x^2))^{-1/2}\), times the derivative of \(\sin(x^2)\) which is \(2x\cos(x^2)\), and the factors of \(2\) cancel to give \(\frac{x\cos(x^2)}{\sqrt{\sin(x^2)}}\). The choice \(\frac{2x\cos(x^2)}{\sqrt{\sin(x^2)}}\) fails to cancel the \(\frac{1}{2}\) from the square root power rule with the \(2\) from the inner quadratic derivative. Multi-layer chain rule problems often simplify nicely once all constant factors are properly combined.

By the chain rule, \(\frac{dy}{dx} = f'(g(x))\cdot g'(x)\), and evaluating at \(x=1\) gives \(f'(g(1))\cdot g'(1) = f'(2)\cdot 4 = 3\cdot 4 = 12\). The choice \(20\) mistakenly multiplies \(f(2)=5\) by \(g'(1)=4\) instead of using \(f'(2)\) as required by the chain rule. When working with table-based chain rule problems, always identify which function value versus derivative value is needed at each stage.

Differentiating implicitly, the left side gives \(-\sin(x-y)(1-\frac{dy}{dx})\) and the right side gives \(\frac{dy}{dx}\sin(x) + y\cos(x)\) using the product rule, and collecting all \(\frac{dy}{dx}\) terms leads to \(\frac{dy}{dx} = \frac{\sin(x-y)+y\cos(x)}{\sin(x)-\sin(x-y)}\). The choice \(\frac{y\cos(x)}{\sin(x)}\) ignores the entire left-hand side contribution from differentiating the composite cosine term. Equations mixing composite trig functions with products of \(x\) and \(y\) require applying both the chain rule and product rule simultaneously before isolating \(\frac{dy}{dx}\).

Using the product rule, the derivative is \(\frac{d}{dx}[e^{x^2}]\cdot\arctan(x) + e^{x^2}\cdot\frac{d}{dx}[\arctan(x)]\), and applying the chain rule to \(e^{x^2}\) gives \(2xe^{x^2}\arctan(x) + \frac{e^{x^2}}{1+x^2}\). The choice \(2xe^{x^2}\arctan(x)\) only accounts for the first term of the product rule and omits the second term entirely. Combining the product rule with the chain rule requires differentiating each factor separately while keeping the other factor unchanged, then adding both resulting terms.

First differentiation gives \(\frac{dy}{dx} = -\frac{y}{x}\), and differentiating again using the quotient rule with \(\frac{dy}{dx}=-\frac{y}{x}\) substituted in gives \(\frac{d^2y}{dx^2} = \frac{2y}{x^2}\) after simplification. The choice \(-\frac{y}{x^2}\) is simply the first derivative divided by \(x\) again rather than the result of properly differentiating the first derivative expression using the quotient rule. Second implicit derivatives require substituting the already-found first derivative back into the differentiated equation before simplifying to a final expression in \(x\) and \(y\).