Consider the function Z(x)=z(x,y(x)), where
z(x,y)=ln(x2+y3)andy(x)=10−x2.
Determine Z′(3).
z(x,y)=ln(x2+y3)andy(x)=10−x2.
Determine Z′(3).
Antwoord 1 correct
Correct
Antwoord 2 optie
Z′(3)=−12.
Antwoord 2 correct
Fout
Antwoord 3 optie
Z′(3)=910.
Antwoord 3 correct
Fout
Antwoord 4 optie
Z′(3)=−133.
Antwoord 4 correct
Fout
Antwoord 1 optie
Z′(3)=−65.
Antwoord 1 feedback
Correct: The partial derivatives of z(x,y) at (x,y(x)) and the derivative of y(x) are:
z′x(x,y)=1x2+y3⋅2x=2xx2+y3z′x(x,y(x))=2xx2+(10−x2)3z′y(x,y)=1x2+y3⋅3y2=3y2x2+y3z′y(x,y(x))=3(10−x2)2x2+(10−x2)3y′(x)=−2x.
According to the Chain rule (case 2):
Z′(x)=2xx2+(10−x2)3+3(10−x2)2x2+(10−x2)3⋅(−2x)=2xx2+(10−x2)3+−6x(10−x2)2x2+(10−x2)3=2x−6x(10−x2)2x2+(10−x2)3.
Finally, we plug in x=3:
Z′(3)=2⋅3−6⋅3(10−32)232+(10−32)3=−1210=−65.
Go on.
z′x(x,y)=1x2+y3⋅2x=2xx2+y3z′x(x,y(x))=2xx2+(10−x2)3z′y(x,y)=1x2+y3⋅3y2=3y2x2+y3z′y(x,y(x))=3(10−x2)2x2+(10−x2)3y′(x)=−2x.
According to the Chain rule (case 2):
Z′(x)=2xx2+(10−x2)3+3(10−x2)2x2+(10−x2)3⋅(−2x)=2xx2+(10−x2)3+−6x(10−x2)2x2+(10−x2)3=2x−6x(10−x2)2x2+(10−x2)3.
Finally, we plug in x=3:
Z′(3)=2⋅3−6⋅3(10−32)232+(10−32)3=−1210=−65.
Go on.
Antwoord 2 feedback
Antwoord 3 feedback
Wrong: Do not forget to apply the chain rule: special case 2.
See Chain rule (case 2), Example 1 and Example 2.
See Chain rule (case 2), Example 1 and Example 2.
Antwoord 4 feedback