{NotebookFileName[],DateString[]}

Human biomass

Population estimate Oct 2017: 7.56 billion

n1=7.6*(10^9)

Natural rate of increase (2010 CIA World Factbook)

r1=Quantity[(8.37/1000),(1/("Years"))]

Mean human body mass, 2005 (Walpole et al.: The weight of nations: an estimation of adult human biomass. BMC Public Health 2012 12:439.)

mbar1=Quantity[62,"Kilograms"]

Rate of increase of US Major League Baseball player body mass, 2.73 lb/10 years (Model 4, Saint Onge, J.M., Krueger, P.M., and Rogers, R.G. (2008). Historical trends in height, weight, and body mass: data from U.S. Major League Baseball players, 1869-1983. Econ. Hum. Biol. 6, 482–488.)

mlbmdot1=Quantity[(2.73/10),(("Pounds")/("Years"))]

Mean mass for the dataset was 181.3 lb.

mlbmbar1=Quantity[181.3,"Pounds"]

I’ll take the ratio to be the relative rate of increase:

mlbmrel1=(mlbmdot1/mlbmbar1)

Rate of population increase

ndot1=n1 r1

mbardot1=mlbmrel1 mbar1

M1=mbar1 n1

Mdot1=n1 mbardot1 + mbar1 ndot1

UnitConvert[Mdot1,(("Kilograms")/("Seconds"))]

UnitConvert[Mdot1,(("Megagrams")/("Minutes"))]

UnitConvert[{n1 mbardot1 , mbar1 ndot1},(("Megagrams")/("Minutes"))]

Quotient rule example: swimming pool

v1[
t_
]:=
4000+100t

m1[
t_
]:=
30*100t

c1[
t_
]:=
((m1[t])/(v1[t]))

D[c1[t],t]

D[c1[t],t]//Simplify

Example of something with infinite limit whose derivative doesn’t have a limit

f2[
x_
]:=
(1/x)Sin[(x^2)]

D[f2[x],x]

Limit[f2[x],x->∞]

Limit[D[f2[x],x],x->∞]

Plot[{
f2[x],
(1/x)
},{x,0,10}
,PlotRange->{All,{-1,1}}
,PlotLegends->{"(1/x)sin((x^2))","(1/x)"}
]

f3[
x_
]:=
Cos[2π Log[x]]

D[f3[x],x]

Limit[D[f3[x],x],x->∞]

Plot[{
f3[x]
},{x,0,10000}
,PlotRange->{All,{-1,1}}
,PlotLegends->{"cos(2π ln(x))"}
]

Inverse trig

D[ArcTan[x],x]

D[ArcSin[x],x]

D[ArcCos[x],x]

Graphs of function and derivative

style[
t_
]:=
Style[t,Bold,FontFamily->"Arial",FontSize->14]

f1[
x_
]:=
(x^3)-3(x^2)+2x+1

fp1[
x_
]:=Evaluate[
D[f1[x],x]
];
fp1[x]

Block[{is=250,a=0,xmin=-1,xmax=3},
Plot[{
f1[x],
fp1[a](x-a)+f1[a]
},{x,xmin,xmax}
,ImageSize->is
,PlotLegends->{"y = f(x)","y = "<>ToString[fp1[a]]<>"·(x-"<>ToString[a]<>") + "<>ToString[f1[a]]}
,Epilog->{
PointSize[Large],Point[{{a,f1[a]}}]
}
]
]

Block[{is=250,a=1,xmin=-1,xmax=3},
Plot[{
f1[x],
fp1[a](x-a)+f1[a]
},{x,xmin,xmax}
,ImageSize->is
,PlotLegends->{"y = f(x)","y = "<>ToString[fp1[a]]<>"·(x-"<>ToString[a]<>") + "<>ToString[f1[a]]}
,Epilog->{
PointSize[Large],Point[{{a,f1[a]}}]
}
]
]

amin=N[Max[a/.Solve[fp1[a]==0,a]]]

Block[{is=250,a=amin,xmin=-1,xmax=3},
Print[Chop[fp1[a]]];
Plot[{
f1[x],
fp1[a](x-a)+f1[a]
},{x,xmin,xmax}
,ImageSize->is
,PlotLegends->{"y = f(x)","y = "<>ToString[Chop[fp1[a]]]<>"·(x-"<>ToString[a]<>") + "<>ToString[f1[a]]}
,Epilog->{
PointSize[Large],Point[{{a,f1[a]}}]
}
]
]

Block[{is=250,a=2.1,xmin=-1,xmax=3},
Print[Chop[fp1[a]]];
Plot[{
f1[x],
fp1[a](x-a)+f1[a]
},{x,xmin,xmax}
,ImageSize->is
,PlotLegends->{"y = f(x)","y = "<>ToString[Chop[fp1[a]]]<>"·(x-"<>ToString[a]<>") + "<>ToString[f1[a]]}
,Epilog->{
PointSize[Large],Point[{{a,f1[a]}}]
}
]
]

Chain rule, linear version

f4[
x_
]:=
Sqrt[((Log[x])^2)+0.02]+0.1(x-(x^2))

Plot[{
f4[x],
f4'[x]
},{x,0,5}
,PlotRange->{All,{-1,2}}
,PlotLegends->{"f(x)","f'(x)"}
]

Plot[{
f4[x]
},{x,0,5}
,PlotRange->{All,{-1,2}}
,PlotLegends->{"f(x)"}
]

f4'[4]//N

Plot[{
f4[x],
f4[2x]
},{x,0,5}
,PlotRange->{All,{-1,2}}
,PlotLegends->{"f(x)","(f(2x))"}
]

Block[{d=0.5,a=4,p1,p11,p12,p2,p21,p22},
p1={a,f4[a]};
p11={a-d,f4[a]-f4'[a]d};
p12={a+d,f4[a]+f4'[a]d};
p2={(a/2),f4[a]};
p21={(a/2)-d,f4[a]-2f4'[a]d};
p22={(a/2)+d,f4[a]+2f4'[a]d};
Plot[{
f4[x],
f4[2x]
},{x,0,5}
,PlotRange->{All,{-1,1}}
,PlotLegends->{"f(x)","(f(2x))"}
,Epilog->{
PointSize[Large],Thickness[0.01],
ColorData[97][1],Line[{p11,p12}],Point[{p1}],
ColorData[97][2],Line[{p21,p22}],Point[{p2}],
Black,Arrow[{p1,p2}]
}
]
]

Block[{d=0.5,a=4,b=-2,p1,p11,p12,p2,p21,p22},
p1={a,f4[a]};
p11={a-d,f4[a]-f4'[a]d};
p12={a+d,f4[a]+f4'[a]d};
p2={((a-b)/2),f4[a]};
p21={((a-b-2d)/2),f4[a]-2f4'[a]d};
p22={((a-b+2d)/2),f4[a]+2f4'[a]d};
Plot[{
f4[x],
f4[2x+b]
},{x,0,5}
,PlotRange->{All,{-1,1}}
,PlotLegends->{"f(x)","(f(2x-2))"}
,Epilog->{
PointSize[Large],Thickness[0.01],
ColorData[97][1],Line[{p11,p12}],Point[{p1}],
ColorData[97][2],Line[{p21,p22}],Point[{p2}],
Black,Arrow[{p1,p2}]
}
]
]

More differentiation problems

PDF[NormalDistribution[μ,σ],x]

f6[
x_
]:=
Exp[-(((x-μ))^2)]

Block[{is=400,μ=5,xmin=0,xmax=10},
Column[{
Plot[{
f6[x],
f6'[x]
},{x,xmin,xmax}
,ImageSize->is
,PlotRange->All
,PlotLegends->TraditionalForm/@{f6[x],f6'[x]}
]
}]
]

f5[
x_
]:=
Exp[Sin[5Sqrt[x]]];
f5[t]//TraditionalForm

D[f5[x],x]

Block[{is=400,xmin=0,xmax=20},
Column[{
Plot[{
f5[x],
f5'[x]
},{x,xmin,xmax}
,ImageSize->is
,PlotLegends->{f5[t],f5'[t]}
]
}]
]

f7[
x_
]:=
(Sqrt[(x^2)+((y[x])^2)]);
f7[x]//TraditionalForm

D[f7[x],x]//Simplify

Harmonic oscillator

hoeq1=m x''[t]==-k x[t]

DSolve[hoeq1,x,t]

Implicit differentiation

f7[
x_
]:=
(Sqrt[(x^2)+((y[x])^2)])

f7[x]

D[f7[x],x]//Simplify

D[f7[x],x]//Simplify//TraditionalForm

Solve[D[f7[x],x]==0,y'[x]]