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Suppose f(x) and g(x) are differentiable functions with the known values:
X: 0,1,2,3,4
f(x): 1,2,1,0,3
g(x): 5,3,4,2,1
f'(x): 0,2,3,6,1
g'(x): 2,3,4,4,5
(please note this is suppose to be a table so the each number is in a row vertically)

Showing calculations, evaluate the following:
a). d/dx [f(x)/x^2] when x=2
b). d/dx [g(f(x))] when x=3
c). d/dx [sqrt g(x)-1] when x=0

Respuesta :

apologiabiology
remember the rules
the deritivive of f(g(x))=f'(g(x))g'(x)
and division thing
the deritivieve of f(x)/g(x)=(f'(x)g(x)-g'(x)f(x))/(g(x)^2)
for ax^n, the deritivive is anx^(n-1)
so

a. f(x)/x^2, deritivive is (f'(x)x^2-2xf(x))/(x^4)
for x=2
(f'(2)(2)^2-2(2)f(2))/(2^4)=(2*4-4*1)/16=(8-4)/16=4/16=1/4

b. g(f(x))=g'(f(x))f'(x) at x=3
g'(f(3))f'(3)=g'(0)(3)=(2)(3)=6

c. (g(x)-1)^(1/2)=(1/2)(g(x)-1)^(-1/2)(g'(x))=[tex] \frac{g'(x)}{2 \sqrt{g(x)-1} } [/tex] at x=0
[tex] \frac{g'(0)}{2 \sqrt{g(0)-1} } [/tex]=[tex] \frac{2}{2 \sqrt{5-1} } [/tex]=[tex] \frac{2}{2 \sqrt{4} } [/tex]=[tex] \frac{2}{2(2) } [/tex]=[tex] \frac{2}{4} [/tex]=1/2

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