Cal help????

Question

I have a few problems that I need help for this quiz I am taking

Find slope of a tangent line function f(x)= x+1/ x² +4 ; x=0

compute the derivative of g(x) = x³ (2+√ x) at x=4

10 pts for answers worked out

Answer 1

1) f(x) = (x + 1)/(x^2 + 4)

Slope at x = 0.

Using the quotient rule,
f'(x) = [ (x^2 + 4) - (x + 1)(2x) ] / (x^2 + 4)^2
f'(x) = [ x^2 + 4 - 2x(x + 1) ] / (x^2 + 4)^2
f'(x) = [ x^2 + 4 - 2x^2 - 2x ] / (x^2 + 4)^2
f'(x) = [ 4 - 2x - x^2 ] / (x^2 + 4)^2

To find the slope of the tangent line at x = 0, solve for f'(0).

f'(0) = [4 - 0 - 0] / (0 + 4)^2
f'(0) = 4/16
f'(0) = 1/4

Therefore, the slope of the tangent line of f(x) at x = 0 is equal to m = 1/4.

Answer 2

Write the first one as
f(x) = x + x^(-2) then
f'(x) = 1 - 2x^(-3)

Write the 2'nd as
g(x) = 2x² + x^(2.5) then
g'(x) = 4x + 2.5x^(1.5)

Both of those rely on the power rule
d/dx x^n = nx^(n-1)

HTH

Doug

Answer 3

f'(x)= 1-2x^(-3)
f'(0)=1

g'(x)=2x^2(2+sqrt(x)) + x^2(0.5x^(-1.5))

g'(5)=129

Answer 4

I forgot how to do this