
a.
To graph: The function f(h)=sin(h)h and estimate limh→0f(h) and then to compare the estimate with π180 and then to state if there is a reason to believe that limit should be π180 .
a.

Answer to Problem 50E
Estimate is equal to π180 .
Explanation of Solution
Given Information: limh→0sin(h)h is to be estimated.
Graph:
Observing the graph, it can inferred that the limit is near about 0.0175 , comparing with π180 , it is found that the estimate is equal to π180 . It is so because π180 is the conversion factor for converting degrees to radians.
b.
To estimate: limh→0cos(h)−1h using graphing utility in degree mode.
b.

Answer to Problem 50E
limh→0cos(h)−1h=0
Explanation of Solution
Given Information: limh→0cos(h)−1h is to be estimated.
Graph:
Observing the graph it can be easily estimated that limh→0cos(h)−1h=0 .
c.
The formula obtained for the derivative of sinx .
c.

Answer to Problem 50E
The formula is ddx(sinx)=π180(cosx) .
Explanation of Solution
Given Information: Degree mode limits are to be used.
Calculation:
ddx(sinx)=limh→0sin(x+h)−sinxh =limh→0sinxcos h+cos xsin h−sin xh =limh→0sinx(cos h−1)+cosxsin hh =(limh→0sinx)(limh→0cos h−1h)+(limh→0cos x)(limh→0sin hh) =(sinx)(0)+(cosx)(π180) =π180(cosx)
Thus, the formula is ddx(sinx)=π180(cosx) .
d.
The formula obtained for the derivative of cosx .
d.

Answer to Problem 50E
The formula is ddx(cosx)=−π180(sinx) .
Explanation of Solution
Given Information: Degree mode limits are to be used.
Calculation:
ddx(cosx)=limh→0cos(x+h)−cosxh =limh→0cosxcos h−sin xsin h−cos xh =limh→0cosx(cos h−1)−sinxsin hh =(limh→0cosx)(limh→0cos h−1h)−(limh→0sin x)(limh→0sin hh) =(cosx)(0)−(sinx)(π180) =−π180(sinx)
Thus, the formula is ddx(cosx)=−π180(sinx) .
e.
To calculate: The second and third degree-mode derivatives of sinx and cosx .
e.

Answer to Problem 50E
The derivatives are d2dx2sinx=−( π180)2sinx, d3dx3sinx =− π31803cosx and d2dx2cosx=−( π180)2cosx, d3dx3cosx=π31803sinx .
Explanation of Solution
Given Information: Degree mode derivatives are to be calculated.
Calculation:
d2dx2sinx=ddx(π180cosx) =π180(−π180sinx) =−( π180)2sinxd3dx3sinx=ddx[−( π180)2sinx] =− π21802(π180cosx) =− π31803cosx
d2dx2cosx=ddx(−π180sinx) =−π180(π180cosx) =−( π180)2cosx
d3dx3cosx=ddx[(− π180)2cosx] =− π21802(−π180sinx) =π31803sinx
Thus, the derivatives are d2dx2sinx=−( π180)2sinx, d3dx3sinx =− π31803cosx and d2dx2cosx=−( π180)2cosx, d3dx3cosx=π31803sinx .
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