Answered

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

Verified Answer :

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)=(24-41)/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))=\(\frac{g'(x)}{2 \sqrt{g(x)-1} }\) at x=0
\(\frac{g'(0)}{2 \sqrt{g(0)-1} }\)=\(\frac{2}{2 \sqrt{5-1} }\)=\(\frac{2}{2 \sqrt{4} }\)=\(\frac{2}{2(2) }\)=\(\frac{2}{4}\)=1/2

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