This comprehensive final exam review covers every major derivative topic — from basic power rules through chain rule, product and quotient rules, to real-world optimization and related rates. Problems are sequenced by section and difficulty. Full step-by-step solutions are included. Ideal for AP Calculus AB/BC final exam preparation, college Calculus I, and comprehensive review.
Use this reference as needed during the review. On your actual exam, these formulas must be memorized.
Power Rule: \(\dfrac{d}{dx}[x^n] = nx^{n-1}\)
Chain Rule: \(\dfrac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x)\)
Product Rule: \(\dfrac{d}{dx}[f\cdot g] = f'g + fg'\)
Quotient Rule: \(\dfrac{d}{dx}\!\left[\dfrac{f}{g}\right] = \dfrac{f'g - fg'}{g^2}\)
| Function | Derivative | Function | Derivative |
|---|---|---|---|
| \(\sin x\) | \(\cos x\) | \(e^x\) | \(e^x\) |
| \(\cos x\) | \(-\sin x\) | \(a^x\) | \(a^x \ln a\) |
| \(\tan x\) | \(\sec^2 x\) | \(\ln x\) | \(1/x\) |
| \(\sec x\) | \(\sec x\tan x\) | \(\log_a x\) | \(1/(x\ln a)\) |
| \(\csc x\) | \(-\csc x\cot x\) | \(\arcsin x\) | \(1/\sqrt{1-x^2}\) |
| \(\cot x\) | \(-\csc^2 x\) | \(\arctan x\) | \(1/(1+x^2)\) |
Apply the power rule, constant multiple rule, sum and difference rules, and basic trig/exponential/log derivatives. No chain, product, or quotient rule needed.
All problems require the chain rule. Some combine it with basic rules. Identify the outer and inner function before differentiating.
Each problem requires the product rule, the quotient rule, or both — often combined with the chain rule. Simplify answers where possible.
This section covers optimization (max/min), related rates, increasing/decreasing intervals, concavity, and linearization. Show all work and include units where applicable.
Apply the power rule to each term: \(\frac{d}{dx}[7x^4]=28x^3\), \(\frac{d}{dx}[-3x^2]=-6x\), \(\frac{d}{dx}[5x]=5\), \(\frac{d}{dx}[-9]=0\).
Rewrite: \(g(x)=x^{4/3}-2x^{-3}\). Differentiate: \(g'(x)=\frac{4}{3}x^{1/3} - 2(-3)x^{-4} = \frac{4}{3}x^{1/3}+6x^{-4}\).
Divide: \(y = x^3 - 4x + 7x^{-2}\). Then \(y'=3x^2 - 4 + 7(-2)x^{-3} = 3x^2-4-\frac{14}{x^3}\).
\(\frac{d}{dt}[6e^t]=6e^t\); \(\frac{d}{dt}[-4\ln t]=-4/t\); \(\frac{d}{dt}[3]=0\).
Term by term: \(\frac{d}{dx}[5\sin x]=5\cos x\); \(\frac{d}{dx}[-2\cos x]=2\sin x\); \(\frac{d}{dx}[\tan x]=\sec^2 x\).
Power rule: \(\frac{d}{dx}[8x^{3/4}]=8\cdot\frac{3}{4}x^{-1/4}=6x^{-1/4}\); \(\frac{d}{dx}[-12x^{1/2}]=-6x^{-1/2}\); \(\frac{d}{dx}[x^{-2}]=-2x^{-3}\).
\(\frac{d}{dx}[3\sec x]=3\sec x\tan x\); \(\frac{d}{dx}[-\cot x]=\csc^2 x\); \(\frac{d}{dx}[2]=0\).
Expand: \((2x-3)^2=4x^2-12x+9\). Differentiate: \(y'=8x-12\).
\(\pi\) and \(e\) are constants. \(f'(x)=3\pi x^2 - e\). (\(\pi^2\) is a constant; its derivative is 0.)
\(\frac{d}{dx}[4\ln x]=4/x\); \(\frac{d}{dx}[7e^x]=7e^x\); \(\frac{d}{dx}[2^x]=2^x\ln 2\).
\(v_0\) and \(s_0\) are constants. \(\frac{ds}{dt}=32t-v_0\). This is the velocity in free-fall kinematics.
\(f'(x)=3x^2-5\). At \(x=2\): \(f'(2)=3(4)-5=7\).
\(\frac{d}{dx}[\arctan x]=\frac{1}{1+x^2}\); \(\frac{d}{dx}[-\arcsin x]=-\frac{1}{\sqrt{1-x^2}}\).
Rewrite: \(f(x)=3x^{-4}+5x^{-1/2}-6x\). Then \(f'(x)=-12x^{-5}-\frac{5}{2}x^{-3/2}-6\).
\(y'=\frac{1}{x}+e^x\). At \(x=1\): point \((1, 0+e)=(1,e)\), slope \(y'(1)=1+e\). Tangent: \(y-e=(1+e)(x-1)\), i.e. \(y=(1+e)x-1\).
Outer: \(u^6\), inner: \(u=3x^2-5x+1\). \(y'=6(3x^2-5x+1)^5\cdot(6x-5)\).
Outer: \(\sin u\), inner: \(u=4x^3-x\). \(f'(x)=\cos(4x^3-x)\cdot(12x^2-1)\).
Outer: \(e^u\), inner: \(u=t^2-3t\). \(g'(t)=e^{t^2-3t}\cdot(2t-3)\).
Outer: \(\ln u\), inner: \(u=x^2+9\). \(y'=\dfrac{1}{x^2+9}\cdot 2x=\dfrac{2x}{x^2+9}\).
Write \(y=(\cos x)^5\). Outer: \(u^5\), inner: \(\cos x\). \(y'=5\cos^4 x\cdot(-\sin x)=-5\cos^4 x\sin x\).
Write \(y=(5x^2-2x+3)^{1/2}\). \(y'=\dfrac{1}{2}(5x^2-2x+3)^{-1/2}\cdot(10x-2)=\dfrac{10x-2}{2\sqrt{5x^2-2x+3}}\).
Outer: \(\tan u\), inner: \(u=e^x\). \(y'=\sec^2(e^x)\cdot e^x=e^x\sec^2(e^x)\).
Outer: \(\ln u\), inner: \(u=\sin t\). \(\dfrac{d}{dt}[\ln(\sin t)]=\dfrac{\cos t}{\sin t}=\cot t\).
Outer: \(e^u\), inner: \(u=\cos x\). \(y'=e^{\cos x}\cdot(-\sin x)=-\sin x\,e^{\cos x}\).
Outer: \(\sec u\), inner: \(u=3x^2\). \(f'(x)=\sec(3x^2)\tan(3x^2)\cdot 6x=6x\sec(3x^2)\tan(3x^2)\).
Outer: \(\arctan u\), inner: \(u=2x^3\). \(y'=\dfrac{1}{1+(2x^3)^2}\cdot 6x^2=\dfrac{6x^2}{1+4x^6}\).
Let \(u=\dfrac{x-1}{x+1}\). By quotient rule: \(u'=\dfrac{(x+1)-(x-1)}{(x+1)^2}=\dfrac{2}{(x+1)^2}\). Then \(y'=4u^3\cdot u'=4\!\left(\dfrac{x-1}{x+1}\right)^3\!\cdot\dfrac{2}{(x+1)^2}=\dfrac{8(x-1)^3}{(x+1)^5}\).
Let \(u=\sec x+\tan x\). \(u'=\sec x\tan x+\sec^2 x=\sec x(\tan x+\sec x)\). So \(y'=\dfrac{\sec x(\sec x+\tan x)}{\sec x+\tan x}=\sec x\).
Let \(v=\sqrt{x^2+1}=(x^2+1)^{1/2}\). Then \(v'=\dfrac{x}{\sqrt{x^2+1}}\). By chain rule on \(e^v\): \(y'=e^{\sqrt{x^2+1}}\cdot\dfrac{x}{\sqrt{x^2+1}}\).
By the Pythagorean identity, \(\sin^2(3x)+\cos^2(3x)=1\) for all \(x\). The derivative of a constant is 0.
Let \(f=x^3, f'=3x^2\); \(g=\sin x, g'=\cos x\). Product rule: \(y'=3x^2\sin x + x^3\cos x\).
\(f=e^x, f'=e^x\); \(g=x^2-4x+1, g'=2x-4\). Product: \(f'(x)=e^x(x^2-4x+1)+e^x(2x-4)=e^x(x^2-2x-3)=e^x(x-3)(x+1)\).
\(f=x^{1/2},f'=\frac{1}{2}x^{-1/2}\); \(g=\ln x, g'=\frac{1}{x}\). Product: \(y'=\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}=\frac{\ln x}{2\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{\ln x+2}{2\sqrt{x}}\).
\(f=t^2,f'=2t\); \(g=e^{-t},g'=-e^{-t}\). Product: \(g'(t)=2te^{-t}-t^2e^{-t}=te^{-t}(2-t)\).
Three-factor product: \((fgh)'=f'gh+fg'h+fgh'\) where \(f=x,g=\cos x,h=\sin x\). \(f'=1,g'=-\sin x,h'=\cos x\). So \(y'=\cos x\sin x + x(-\sin x)\sin x + x\cos x\cos x = \cos x\sin x - x\sin^2 x + x\cos^2 x\). Simplify: \(=\cos x\sin x + x(\cos^2 x-\sin^2 x)=\frac{1}{2}\sin 2x + x\cos 2x\).
Quotient: \(f=x^2+3x,f'=2x+3\); \(g=x-5,g'=1\). \(y'=\dfrac{(2x+3)(x-5)-(x^2+3x)}{(x-5)^2}=\dfrac{2x^2-7x-15-x^2-3x}{(x-5)^2}=\dfrac{x^2-10x-15}{(x-5)^2}\).
\(f=\sin x,f'=\cos x\); \(g=x^2,g'=2x\). Quotient: \(y'=\dfrac{x^2\cos x-2x\sin x}{x^4}=\dfrac{x\cos x-2\sin x}{x^3}\).
\(f=e^x,f'=e^x\); \(g=x^3+1,g'=3x^2\). Quotient: \(h'(x)=\dfrac{e^x(x^3+1)-e^x(3x^2)}{(x^3+1)^2}=\dfrac{e^x(x^3-3x^2+1)}{(x^3+1)^2}\).
\(f=\ln x,f'=1/x\); \(g=x^{1/2},g'=\frac{1}{2}x^{-1/2}\). Quotient: \(y'=\dfrac{\frac{1}{x}\cdot\sqrt{x}-\ln x\cdot\frac{1}{2\sqrt{x}}}{x}=\dfrac{\frac{1}{\sqrt{x}}-\frac{\ln x}{2\sqrt{x}}}{x}=\dfrac{2-\ln x}{2x^{3/2}}\).
\(f=x^2-1,f'=2x\); \(g=x^2+1,g'=2x\). Quotient: \(f'(x)=\dfrac{2x(x^2+1)-2x(x^2-1)}{(x^2+1)^2}=\dfrac{2x(x^2+1-x^2+1)}{(x^2+1)^2}=\dfrac{4x}{(x^2+1)^2}\).
\(f=x^2,f'=2x\); \(g=\arctan x,g'=\frac{1}{1+x^2}\). Product: \(y'=2x\arctan x+\dfrac{x^2}{1+x^2}\).
\(f'(x)=e^x\cos x+e^x(-\sin x)=e^x(\cos x-\sin x)\). For \(f''(x)\): product rule again: \(f''(x)=e^x(\cos x-\sin x)+e^x(-\sin x-\cos x)=e^x(\cos x-\sin x-\sin x-\cos x)=-2e^x\sin x\).
\(f=\tan x,f'=\sec^2 x\); \(g=1+x^2,g'=2x\). Quotient: \(y'=\dfrac{\sec^2 x(1+x^2)-2x\tan x}{(1+x^2)^2}\).
Product rule: \(p'(x)=f'(x)g(x)+f(x)g'(x)\). At \(x=2\): \(p'(2)=(-1)(5)+(3)(4)=-5+12=7\).
Quotient rule: \(q'(x)=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\). At \(x=2\): \(q'(2)=\dfrac{(-1)(5)-(3)(4)}{5^2}=\dfrac{-5-12}{25}=\dfrac{-17}{25}\).
\(f'(x)=3x^2-12x+9=3(x^2-4x+3)=3(x-1)(x-3)\). Critical points: \(x=1\) and \(x=3\). Sign of \(f'\): positive on \((-\infty,1)\), negative on \((1,3)\), positive on \((3,\infty)\).
\(f'(x)=6x^2-18x+12=6(x-1)(x-2)\). Critical points: \(x=1,x=2\). \(f''(x)=12x-18\). \(f''(1)=-6<0\) → local max; \(f''(2)=6>0\) → local min. Values: \(f(1)=2-9+12+1=6\); \(f(2)=16-36+24+1=5\).
\(f'(x)=4x^3-12x^2+8x=4x(x-1)(x-2)\). Critical points in \([-1,3]\): \(x=0,1,2\). Evaluate: \(f(-1)=1+4+4=9\); \(f(0)=0\); \(f(1)=1-4+4=1\); \(f(2)=16-32+16=0\); \(f(3)=81-108+36=9\).
\(f''(x)=12x^2-16\). Set \(f''(x)=0\): \(x^2=\frac{4}{3}\Rightarrow x=\pm\frac{2}{\sqrt{3}}\). Check sign changes: \(f''<0\) on \(\!\left(-\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}\right)\) (concave down), \(f''>0\) outside (concave up). Inflection points at \(x=\pm\frac{2\sqrt{3}}{3}\).
\(f'(x)=\frac{1}{2\sqrt{x}}\). At \(a=25\): \(f(25)=5\), \(f'(25)=\frac{1}{10}\). \(L(x)=5+\frac{1}{10}(x-25)\). Estimate: \(L(26)=5+\frac{1}{10}(1)=5.1\). (True value: \(\sqrt{26}\approx5.099\).)
Let width \(= x\) (two sides), length \(= y\) (one side parallel to river). Constraint: \(2x+y=200\Rightarrow y=200-2x\). Area: \(A=xy=x(200-2x)=200x-2x^2\). \(\frac{dA}{dx}=200-4x=0\Rightarrow x=50\,\text{m}\), \(y=100\,\text{m}\). Second derivative \(-4<0\) confirms maximum.
Let numbers be \(x\) and \(50-x\). Product \(P=x(50-x)=50x-x^2\). \(P'=50-2x=0\Rightarrow x=25\). Both numbers are 25; \(P''=-2<0\) confirms maximum.
Let side of square base \(=x\), height \(=h\). Volume: \(x^2 h=32\Rightarrow h=32/x^2\). Surface area (open top): \(S=x^2+4xh=x^2+\frac{128}{x}\). \(\frac{dS}{dx}=2x-\frac{128}{x^2}=0\Rightarrow x^3=64\Rightarrow x=4\,\text{m}\). Then \(h=32/16=2\,\text{m}\).
Let the foot of the ladder be at distance \(x\) from the wall, so the height is \(y=\sqrt{100-x^2}\). The area of the triangle is \(A=\frac{1}{2}xy=\frac{1}{2}x\sqrt{100-x^2}\). Set \(\frac{dA}{dx}=0\): \(\frac{dA}{dx}=\frac{1}{2}\!\left[\sqrt{100-x^2}+x\cdot\frac{-x}{\sqrt{100-x^2}}\right]=\frac{100-2x^2}{2\sqrt{100-x^2}}=0\Rightarrow x^2=50\Rightarrow x=5\sqrt{2}\,\text{m}\). Height \(=5\sqrt{2}\,\text{m}\).
\(V=\frac{4}{3}\pi r^3\Rightarrow\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\). At \(r=5,\frac{dr}{dt}=2\): \(\frac{dV}{dt}=4\pi(25)(2)=200\pi\,\text{cm}^3/\text{s}\).
Pythagorean relation: \(x^2+y^2=100\). Differentiate: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\Rightarrow\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}\). When \(x=6\): \(y=\sqrt{100-36}=8\). With \(\frac{dx}{dt}=2\): \(\frac{dy}{dt}=-\frac{6}{8}(2)=-\frac{3}{2}\,\text{ft/s}\).
Since \(r/h=4/8=1/2\), we have \(r=h/2\). Volume: \(V=\frac{1}{3}\pi\!\left(\frac{h}{2}\right)^2 h=\frac{\pi h^3}{12}\). Differentiate: \(\frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}\). With \(\frac{dV}{dt}=-2\) (draining) and \(h=4\): \(-2=\frac{\pi(16)}{4}\frac{dh}{dt}\Rightarrow\frac{dh}{dt}=-\frac{2}{4\pi}=-\frac{1}{2\pi}\,\text{m/min}\).
Let \(z\) be distance between cars. \(z^2=x^2+y^2\) where \(x=80t,y=60t\). After 1 hour: \(x=80,y=60,z=100\). \(2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\Rightarrow\frac{dz}{dt}=\frac{80(80)+60(60)}{100}=\frac{6400+3600}{100}=100\,\text{mph}\).
\(f\) is continuous on \([0,2]\) (polynomial) and differentiable on \((0,2)\) — MVT applies. Average rate: \(\frac{f(2)-f(0)}{2-0}=\frac{(8-2)-0}{2}=3\). Set \(f'(c)=3x^2-1=3\Rightarrow x^2=\frac{4}{3}\Rightarrow c=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\approx1.155\in(0,2)\). ✓
\(P'(x)=-0.04x+120\). At \(x=2000\): \(P'(2000)=-0.04(2000)+120=-80+120=40\) dollars/unit. Since \(P'(2000)>0\), profit is still increasing — production should be increased. At \(x=3000\), \(P'=0\) (maximum profit).
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