\(y = x^2 - 3x + 4; x = 7\)

\(f(x) = x^2 - 3x + 4\)

\(f(7+h) = (7+h)^2 - 3(7+h) + 4\)

\(= 49 + h^2 + 14h - 21 - 3h + 4\)

\(= h^2 + 11h + 32\)

\(f(7) = 7^2 - 3.7 + 4\)

\(= 49 - 21 + 4 = 32\)

\(f(7+h) - f(7) = h^2 + 11h + 32 - 32\)

\(= h^2 + 11h\)

\(\lim_{h \to 0} \dfrac{f(7+h) - f(7)}{h} = \lim_{h \to 0} \dfrac{h^2+11h}{h}\)

\(= \lim_{h \to 0} h + 11 = 11\)

\(y = \dfrac{1}{2x+3}; x=-1\)

\(f(x) = \dfrac{1}{2x+3}\)

\(f(-1 + h) = \dfrac{1}{2(h-1)+3} = \dfrac{1}{2h-2+3}=\dfrac{1}{2h+1}\)

\(f(-1) = \dfrac{1}{-2+3} = 1\)

\(\lim_{h \to 0} \dfrac{f(-1+h)-f(-1)}{h}\)

\(= \lim_{h \to 0} \dfrac{\dfrac{1}{2h+1}-1}{h}\)

\(= \lim_{h \to 0} \dfrac{1-2h-1}{h(1+2h)}\)

\(= \lim_{h \to 0} \dfrac{-2h}{h(1+2h)}\)

\(= \lim_{h \to 0} \dfrac{-2}{1+2h}\)

\(= \lim_{h \to 0} \dfrac{-2}{1+0} = -2\)

\(y = x^3\)

\(f(x) = x^3\)

\(f(2+h) = (2+h)^3 = 2^3 + h^3 + 12h + 6h^2\)

\(f(2) = 2^3 = 8\)

\(f(2+h) - f(2) = h^3 + 12h + 6h^2\)

\(\lim_{h \to 0} \dfrac{f(2+h)-f(2)}{h}\)

\(= \lim_{h \to 0} h^2 + 12 + 6h\)

\(= 0 + 12 + 0\)

\(12\)

\(y = \sqrt{5-2x}\)

\(f(x) = \sqrt{5-2x}\)

\(f(-2+h) = \sqrt{5-2(-2+h)}\)

\(= \sqrt{5+4-2h} = \sqrt{9-2h}\)

\(f(-2) = \sqrt{5+4} = \sqrt{9} = 3\)

\(f(-2+h)-f(-2)=\sqrt{9-2h}+3\)

\(\lim_{h \to 0} \dfrac{f(-2+h)-f(-2)}{h}\)

\(= \lim_{h \to 0} \dfrac{\sqrt{9-2h}+3}{h} * \dfrac{\sqrt{9-2h}-3}{\sqrt{9-2h}-3}\)

\(= \lim_{h \to 0} \dfrac{9-2h-9}{h(\sqrt{9-2h}-3)}\)

\(= \lim_{h \to 0} \dfrac{-2}{\sqrt{9-2h}-3}\)

\(= \dfrac{-2}{-3-3}\)

\(= \dfrac{-2}{-6} = \dfrac{1}{3}\)

\(y = \dfrac{1}{x^2}\)

\(f(x) = \dfrac{1}{x^2}\)

\(f(4+h) = \dfrac{1}{(4+h)^2}\)

\(f(4) = \dfrac{1}{16}\)

\(f(4+h) - f(4) = \dfrac{1}{(4+h)^2} - \dfrac{1}{16}\)

\(= \dfrac{16 - (16 + h^2 + 8h)}{16(4+h)^2}\)

\(= \dfrac{-h^2 - 8h}{16(4+h)^2} = \dfrac{-h(h+8)}{16(4+h)^2}\)

\(\lim_{h \to 0} \dfrac{f(4+h)-f(4)}{h}\)

\(= \lim_{h \to 0} \dfrac{-(h+8)}{16(4+h)^2}\)

\(= \dfrac{-8}{16.4^2} = \dfrac{-1}{32}\)

\(y = \sin x\)

\(f(x) = \sin x\)

\(f(0 + h) = \sin (0 + h) = \sin h\)

\(f(0) = \sin 0 = 0\)

\(\lim_{h \to 0} \dfrac{f(0+h)-f(0)}{h} = \lim_{h \to 0} \dfrac{\sin h}{h} = 1\) (Exercise 23, 2.7 section)

\(y = \sin x; x = \dfrac{\pi}{3}\)

\(f(x)=\sin x\)

\(f(\dfrac{\pi}{3}+h) = \sin(\dfrac{\pi}{3}+h)\)

\(f(\dfrac{\pi}{3}) = \sin(\dfrac{\pi}{3})\)

\(\sin(\dfrac{\pi}{3}+h) = \sin (\dfrac{\pi}{3}) \cos h + \cos (\dfrac{\pi}{3}) \sin h\)

\(\sin(\dfrac{\pi}{3}+h) - \sin (\dfrac{\pi}{3}) = \cos(\dfrac{\pi}{3})\sin h + \sin (\dfrac{\pi}{3})(\cos h - 1)\)

\(= \cos(\dfrac{\pi}{3})\sin h + \sin (\dfrac{\pi}{3}) \dfrac{(\cos h - 1)}{(\cos h + 1)} (\cos h + 1)\)

\(= \cos(\dfrac{\pi}{3})\sin h + \sin (\dfrac{\pi}{3}) \dfrac{\cos^2 h - 1}{\cos h + 1}\)

\(= \cos(\dfrac{\pi}{3})\sin h + \sin (\dfrac{\pi}{3}) \dfrac{\sin^2 h}{\cos h + 1}\)

\(= \sin h (\cos (\dfrac{\pi}{3}) - \sin (\dfrac{\pi}{3}) \dfrac{\sin h}{\cos h + 1})\)

\(\lim_{h \to 0} \dfrac{\sin h}{h} (\cos (\dfrac{\pi}{3}) - \sin (\dfrac{\pi}{3}) \dfrac{\sin h}{\cos h + 1})\)

\(= \cos (\dfrac{\pi}{3}) - 0 = \cos (\dfrac{\pi}{3}) = \dfrac{1}{2}\)

\(y = \tan 3x\)

\(f(x) = \tan 3x\)

\(f(0+h) = \tan 3(0+h) = \tan 3h\)

\(f(0) = \tan 0 = 0\)

\(\lim_{h \to 0} \dfrac{\tan 3h}{h}\)

\(= \lim_{h \to 0} \dfrac{3 \tan 3h}{3h}\)

\(= \lim_{h \to 0} \dfrac{3 \sin 3h}{3h(\cos 3h)}\)

\(= 3 . \dfrac{1}{\cos 0} = 3\)

\(y = \dfrac{1}{\sqrt{x}}\)

\(f(2+h) = \dfrac{1}{\sqrt{2+h}}\)

\(f(2) = \dfrac{1}{\sqrt{2}}\)

\(f(2+h)-f(2) = \dfrac{1}{\sqrt{2+h}} - \dfrac{1}{\sqrt{2}}\)

\(= \dfrac{\sqrt{2}-\sqrt{2+h}}{\sqrt{2+h}\sqrt{2}} * \dfrac{\sqrt{2}+\sqrt{2+h}}{\sqrt{2}+\sqrt{2+h}}\)

\(= \dfrac{2-(2+h)}{(\sqrt{2+h}\sqrt{2})(\sqrt{2}+\sqrt{2+h})}\)

\(= \dfrac{-h}{(\sqrt{2+h}\sqrt{2})(\sqrt{2}+\sqrt{2+h})}\)

\(\lim_{h \to 0} \dfrac{-h}{h(\sqrt{2+h}\sqrt{2})(\sqrt{2}+\sqrt{2+h})}\)

\(= \dfrac{-1}{\sqrt{2}\sqrt{2}(\sqrt{2} + \sqrt{2})}\)

\(= \dfrac{-1}{2(2\sqrt{2})} = \dfrac{-1}{4\sqrt{2}} = -\dfrac{\sqrt{2}}{8}\)

\(y = 3x^2 - 5x + 1\)

\((x,y) = (2,3)\)

Let's find the slope of the line tangent to the graph \(x=3\).

We know that the instantaneous rate of change of \(y\) with respect to \(x\) is that tangent at \(x=2\).

\(f(2+h) = 3(2+h)^2 - 5(2+h) + 1\)

\(= 3(4+h^2 + 4h) - 10 - 5h + 1\)

\(= 12 + 3h^2 + 12h - 10 - 5h + 1\)

\(= 3 + 3h^2 + 7h\)

\(f(2) = 3.2^2 - 5.2 + 1\)

\(= 12 - 10 + 1 = 3\)

\(f(2+h) - f(2) = 3 + 3h^2 + 7h - 3\)

\(= 3h^2 + 7h\)

\(\lim_{h \to 0} \dfrac{f(2+h)-f(2)}{h} = \lim_{h \to 0} 3h + 7 = 7\)

So, the slope is 7.

Using the equation of line \(y = mx + c\)

\(3 = 7.2 + c\)

\(c = 3 - 14 = 11\)

\(y = mx + c\)

\(y = 7x - 11\)

We know that the instantaneous rate of change of \(y\) with respect to \(x\) at \(x=a\) is the tangent at \(x=a\).

\(y = \dfrac{1}{x^2}\)

\((-1,1)\)

\(f(x) = \dfrac{1}{x^2}\)

\(f(-1 + h) = \dfrac{1}{(h-1)^2} = \dfrac{1}{h^2 + 1 - 2h}\)

\(f(-1) = 1\)

\(f(-1 + h) - f(-1) = \dfrac{1}{h^2+1-2h} - 1 = \dfrac{1-(h^2+1-2h)}{h^2+1-2h}\)

\(= \dfrac{-h^2 + 2h}{h^2+1-2h}\)

\(\lim_{h \to 0} \dfrac{f(-1+h)-f(-1)}{h} = \lim_{h \to 0} \dfrac{-h+2}{h^2 + 1 -2h}\)

\(= \dfrac{2}{1} = 2\)

So, \(m = 2\)

\(y = mx + c\)

\(1 = 2.-1 + c\)

\(c = 1 + 2 = 3\)

\(y = mx + c\)

\(y = 2x + 5\)

\(y = t + \sqrt{t}\)

\(t = time in seconds\)

\(y = distance in inches\)

\(I = \dfrac{10}{R}\)

\(R\) = Resistance in Ohms.

\(I\) = Current in ampere.

Rate of change of \(I\) with respect to \(R\) when \(R = 5 ohms\).

\(f(R) = \dfrac{10}{R}\)

\(f(5+h) = \dfrac{10}{5+h}\)

\(f(5) = \dfrac{10}{5} = 2\)

\(f(5+h) - f(5) = \dfrac{10}{5+h} - 2 = \dfrac{10-10-2h}{5+h}\)

\(= \dfrac{-2h}{5+h}\)

\(\lim_{h \to 0} \dfrac{f(5+h)-f(5)}{h} = \lim_{h \to 0} \dfrac{-2}{5+h}\)

\(= \dfrac{-2}{5}\)

So, the instantaneous rate of change of \(I\) with respect to \(R\) is \(-\dfrac{2}{5}\) ampere per ohm.

Since it's negative, it's decreasing. So as the resistance increases, current decreases.

\(F = \dfrac{k}{r^2}\)

\(k \approx 23.07\)

\(F =\) force in degrees.

\(r =\) Distance in picometers.

Rate of change of \(F\) with respect to \(r\) when \(r= 30\) pm.

\(f(r) = \dfrac{k}{r^w}\)

\(f(30+h) = \dfrac{k}{(30+h)^2} = \dfrac{k}{30^2+h^2+60h}\)

\(f(30) = \dfrac{k}{30^2}\)

\(f(30+h)-f(30) = k(\dfrac{1}{30^2+h^2+60h} - \dfrac{1}{30^2})\)

\(= k(\dfrac{30^2-30^2-h^2-60h}{(30^2)(30^2+h^2+60h)})\)

\(= k(\dfrac{-h^2-60h}{(30^2)(30^2+h^2+60h)})\)

\(\lim_{h \to 0} \dfrac{f(30+h)-f(30)}{h}\)

\(= \lim_{h \to 0} \dfrac{k(-h-60)}{30^2(30^2+h^2+60h)}\)

\(= \dfrac{k(-60)}{30^2(30^2 + 0 + 6)}\)

\(= \dfrac{-60k}{30^2(30^2)}\)

\(= - 0.0017\) dynes per pm

A negative rate of change indicates that as distance increases, the force decreases.

\(P = r^{\dfrac{3}{2}}\)

P = Time to complete one orbit in years.

r = Circular orbit of radius r in astronomical units.

Rate of change of \(P\) with respect to \(r\) at \(r=4 au\)

\(f(r) = r^{3/2}\)

\(f(4+h) = (4+h)^{3/2}\)

\(f(4) = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8\)

\(f(4+h) - f(4) = \sqrt{(4+h)^3} - 8\)

\(= \sqrt{(4+h)^3} - 8 * \dfrac{\sqrt{(4+h)^3} + 8}{\sqrt{(4+h)^3} + 8}\)

\(= \dfrac{(4+h)^3 - 64}{\sqrt{(4+h)^3}+8}\)

\(= \dfrac{64+h^3+3.4^2h+3.4.h^2-64}{\sqrt{(4+h)^3}+8}\)

\(= \dfrac{h^3+3.4^2.h+3.4.h^2}{\sqrt{(4+h)^3}+8}\)

\(\lim_{h \to 0} \dfrac{f(4+h)-f(4)}{h}\)

\(= \lim_{h \to 0} \dfrac{h^2+3.4^2+34.h}{\sqrt{(4+h)^3}+8}\)

\(= \dfrac{0+3.4^2 + 0}{\sqrt{4^3}+8}\)

\(= \dfrac{3.4^2}{2^3+8}\)

\(= \dfrac{3.4.4}{8+8}\)

\(= \dfrac{3.2}{1+1}\)

\(= 3\)

So, the rate of change of \(P\) with respect to \(r\) when \(r=4\) au is 3 years per au.

Since it is positive, \(P\) increases as \(r\) increases.

\(L = kT^4\)

\(k \approx 3.45 * 10^{11}\)

T = Surface temperature in Kelvin

L = Luminosity measured in watts

Rate of change of \(L\) with respect to \(T\) when \(T=2000 K\).

\(f(T) = kT^4\)

\(f(2000+h) = k(2000+h)^4\)

\(= k(2000^2+h^2+4000h)(2000+h)^2\)

\(= k(2000^2+h^2+4000h)(2000^2+h^2+4000h)\)

\(= k(2000^4 + 2000^2h^2 + 2000^2.4000.h+2000^2h^2 + h^4 + 4000h^3 + 2000^2.4000h + 4000h^3 + 4000^2.h^2)\)

\(f(2000) = k(2000)^4\)

\(f(2000+h)-f(2000) = k(2000^2h^2 + 2000^2.4000.h+2000^2h^2 + h^4 + 4000h^3 + 2000^2.4000h + 4000h^3 + 4000^2.h^2)\)

\(= k(h^2(2000^2 + 2000^2 + 4000^2) + h(2000^2.4000 + 2000^2.4000) + h^4 + h^3(4000 + 4000))\)

\(\lim_{h \to 0} \dfrac{f(2000+h)-f(2000)}{h}\)

\(= \lim_{h \to 0} k(h(2000^2 + 2000^2 + 4000^2) + (2000^2.4000 + 2000^2.4000) + h^3 + h^2(4000 + 4000))\)

\(= k(2*2000^2.4000)\)

\(= 8000(2000)^2.k\)

\(= 30 * 10^9\) K

So, the rate of change of \(L\) with respect to \(T\) when \(T=2000\) K is \(32*10^9\) watts per kelvin.