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  5. Derivative
  6. Exercise 2

Exercise 2

Determine the derivative of y(x)=−2x2−5x−1 in x=−1.
y′(−1)=−1.
y′(−1)=−5
y′(−1)=1.
y′(−1) does not exist, because the difference quotient goes to infinity if Δx goes to 0.
Determine the derivative of y(x)=−2x2−5x−1 in x=−1.
Antwoord 1 correct
Correct
Antwoord 2 optie
y′(−1)=−5
Antwoord 2 correct
Fout
Antwoord 3 optie
y′(−1)=1.
Antwoord 3 correct
Fout
Antwoord 4 optie
y′(−1) does not exist, because the difference quotient goes to infinity if Δx goes to 0.
Antwoord 4 correct
Fout
Antwoord 1 optie
y′(−1)=−1.
Antwoord 1 feedback
Correct: For the difference quotient with start value x=−1 and Δx we need y(−1) and y(−1+Δx):
y(2)=−2⋅(−1)2−5⋅(−1)−1=2,y(−1+Δx)=−2(−1+Δx)2−5(−1+Δx)−1=−2(1−2Δx+(Δx)2)−5(−1+Δx)−1=−2+4Δx−2(Δx)2+5−5Δx−1=−2(Δx)2−Δx+2.
Plugging these values into the difference quotient gives
ΔyΔx=y(−1+Δx)−y(−1)Δx=−2(Δx)2−Δx+2−2Δx=−2(Δx)2−ΔxΔx=−2Δx−1.
If Δx→0, then ΔyΔx→−1, hence y′(−1)=−1.

Go on.
Antwoord 2 feedback
Wrong: Be careful when working out brackets. (−1+Δx)2≠1+(Δx)2.

See also Example.
Antwoord 3 feedback
Wrong: Pay attention to the order of y(−1) and y(−1+Δx) in the nominator of the difference quotient.

See also Difference quotient and Example.
Antwoord 4 feedback
Wrong: Be careful when working out brackets and when dealing with minus-signs. You probably made a mistake rewriting.

See also Example.