Derivative at a Point Worksheet

Corresponding to Section 1.3

Given that k(x)=(1.5^x), we want the derivative at x=-2.

Graph this function along with its tangent line at x=-2. Approximate the slope of this tangent line.

g1=Graphics[{Thickness[0.004],Line[{{-10,-10},{-10,10}}],Line[{{-5,-10},{-5,10}}],Line[{{5,-10},{5,10}}],Line[{{10,-10},{10,10}}],Line[{{-10,-10},{10,-10}}],Line[{{-10,-5},{10,-5}}],Line[{{-10,5},{10,5}}],Line[{{-10,10},{10,10}}]},Axes->True,PlotRange->{{-10,10},{-2,10}},GridLines->{Range[-10,10],Range[-2,10]},AxesStyle->Thick];
k[x_]:=1.5^x
Show[{g1,Plot[k[x],{x,-10,10}]},Plot[k'[-2](x-(-2))+k[-2],{x,-10,10},PlotStyle->Red],Plot[k'[2](x-2)+k[2],{x,-10,10},PlotStyle->Yellow]]

{k'[-2],2.6/15}

Approximate the derivative of this function at x=2 to 3 decimal places by numerically evaluating the difference quotient for x-values that get progressively closer to 2.
	({{x, 2.1, 2.01, 2.001, 2.0001, 2.00001}, {((k(x)-k(2))/(x-2)),                          ,                      ,                      ,                      ,                      }})

	k'(2)≈________________

TableForm[
Transpose[
Table[{x,(k[x]-k[2])/(x-2)},{x,{2.1,2.01,2.001,2.0001,2.00001}}]
],TableHeadings->{{"x","((k (x) - k (2))/(x - 2))"},{}}
]
Print["k'(2) is approximately ", 1.0*Floor[1000*k'[2]]/1000]

{k'[2],9.0/10.0}

Given that f(x)=3x-2, we want the derivative at x=1.

Graph this function along with its tangent line at x=1. Identify the slope of this tangent line.

g2=Graphics[{Thickness[0.004],Line[{{-10,-10},{-10,10}}],Line[{{-5,-10},{-5,10}}],Line[{{5,-10},{5,10}}],Line[{{10,-10},{10,10}}],Line[{{-10,-10},{10,-10}}],Line[{{-10,-5},{10,-5}}],Line[{{-10,5},{10,5}}],Line[{{-10,10},{10,10}}]},Axes->True,PlotRange->{{-10,10},{-10,10}},GridLines->{Range[-10,10],Range[-10,10]},AxesStyle->Thick];
f[x_]:=3x-2
Show[{
g2,
Plot[f[x],{x,-10,10},PlotStyle->{Blue,Thick}],Plot[f'[1](x-(1))+f[1],{x,-10,10},PlotStyle->{Red,Dashed}]}
, PlotLabel->"f'[1]=3"
]

Find the derivative of this function at x=1 using the limit definition.

f’(x)
 = (lim_(h->0))(f(1+h)-f(1))/h

= (lim_(h->0))(3*(1+h)-2-(3*1-2))/h

= (lim_(h->0))(3h)/h = 
 (lim_(h->0))3 = 3

Given that g(x)=((x^2)/2), we want the derivative at x=3.

Graph this function along with its tangent line at x=3. Identify the slope of this tangent line.

g2=Graphics[{Thickness[0.004],Line[{{-10,-10},{-10,10}}],Line[{{-5,-10},{-5,10}}],Line[{{5,-10},{5,10}}],Line[{{10,-10},{10,10}}],Line[{{-10,-10},{10,-10}}],Line[{{-10,-5},{10,-5}}],Line[{{-10,5},{10,5}}],Line[{{-10,10},{10,10}}]},Axes->True,PlotRange->{{-10,10},{-10,10}},GridLines->{Range[-10,10],Range[-10,10]},AxesStyle->Thick];
g[x_]:=x^2/2
x0=3;
Show[{
g2,
Plot[g[x],{x,-10,10},PlotStyle->{Blue,Thick}],Plot[g'[x0](x-x0)+g[x0],{x,-10,10},PlotStyle->{Red,Dashed}]}
, PlotLabel->"g'[3]=3"
]

Find the derivative of this function at x=3 using the limit definition.

g’(x)
 = (lim_(h->0))(g(3+h)-g(3))/h

= (lim_(h->0))((3+h)^2/2-(3^2/2))/h

= (lim_(h->0))((6h+h^2)/2)/h = 
 (lim_(h->0))(3+h/2) = 3

Given that h(x)=(x^2)-x, we want the derivative at x=2.

Graph this function along with its tangent line at x=2. Identify the slope of this tangent line.

g2=Graphics[{Thickness[0.004],Line[{{-10,-10},{-10,10}}],Line[{{-5,-10},{-5,10}}],Line[{{5,-10},{5,10}}],Line[{{10,-10},{10,10}}],Line[{{-10,-10},{10,-10}}],Line[{{-10,-5},{10,-5}}],Line[{{-10,5},{10,5}}],Line[{{-10,10},{10,10}}]},Axes->True,PlotRange->{{-10,10},{-10,10}},GridLines->{Range[-10,10],Range[-10,10]},AxesStyle->Thick];
g[x_]:=x^2-x
x0=2;
Show[{
g2,
Plot[g[x],{x,-10,10},PlotStyle->{Blue,Thick}],Plot[g'[x0](x-x0)+g[x0],{x,-10,10},PlotStyle->{Red,Dashed}]}
, PlotLabel->"g'[2]=3"
]

Find the derivative of this function at x=2 using the limit definition.

It’s a little confusing to use “h” for the function name when you’re doing derivatives by the limit definition; hence I’ll use “g”:

g’(x)
 = (lim_(h->0))(g(2+h)-g(2))/h

= (lim_(h->0))((2+h)^2-(2+h)-(2^2-2))/h

= (lim_(h->0))((4h+h^2)-h)/h = 
 (lim_(h->0))(3) = 3

Find an equation for the tangent line at x=2.

Print["y=",g'[2](x-2)+g[2]]