Let F(x)=((x^2)+4)^5 Find The Differential Dx When

MrMan4

From: US
Posts: 8
Votes: 32
Let f(x)=((x^2)+4)^5 Find the differential dx when x=4 and dx=0.4 Find the differential dx when x=4 and dx=0.05

 

Answer No.1

WizTeacher

From: US
Posts: 18
Votes: 54
F(x)=((x^2)+4)^5 when x=4 dx=0.4 f'(x) = 5((x^2)+4)^4 * (2x)dx = 10x((x^2)+4)^4dx = 10(4)(16+4)^4 * 0.4 = 2560000 -> ANSWER when x=4 dx = 0.05 =10(4)(16+4)^4 * 0.0.5 = 320000

Answer No.2

ProBrainy9

From: US
Posts: 13
Votes: 26
F(x)=(x2+4)5 When x=4 and dx=0.4, f'(x) = 5(x2+4)4 (2x)dx = 10x(x2+4)4dx f'(x) = 10(4)(16+4)4(0.4) =40*160000*.4= 2560000 When x=4 and dx = 0.05, f'(x)=10(4)(16+4)4 (.5) = 320000

Answer No.3

Anonymous

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