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How to find derivative at a point?

0 votes
How do I find a derivative at a point for the following two functions?

f(x) = sinx at pi/4
g(x) = cosx at pi/4
asked May 11, 2013 in TRIGONOMETRY by abstain12 Apprentice

4 Answers

0 votes

Given f(x) = sinx 

Differentiate each side with respective x

f'(x) = d/dx(sinx)

Recall differentiation formulas : d/dx(sinx) = cosx

f'(x) = cosx

Derivative at a point pi/4 is

f'(pi/4) = cos(pi/4)

Recall trigonometric tabular form values : cos(pi/4) = 1/sqrt(2)

f'(pi/4) = 1/sqrt(2)

f'(pi/4) = 0.707

 

answered May 13, 2013 by jeevitha Novice
0 votes

Given g(x) = cosx

Differentiate each side with respective x

g'(x) = d/dx(cosx)

Recall differentiation formulas : d/dx(cosx) = -sinx

g'(x) = -sinx

Derivative at a point pi/4 is

g'(pi/4) = -sin(pi/4)

Recall trigonometric tabular form values : sin(pi/4) = 1/sqrt(2)

g'(pi/4) = -1/sqrt(2)

g'(pi/4) = -0.707

Derivative at a point for the function g(x) = cosx at pi/4 = -0.707

answered May 13, 2013 by jeevitha Novice
0 votes

Derivative

f(x) = sinx at x = π / 4

f(π / 4) = sin(π / 4) = 1 / √(2)

The point of the function = (π / 4 , 1 / √(2))

f(x) = sinx

Derivative with respective x

f '(x) = cosx at point (π / 4 , 1 / √(2))

f ' (π / 4) = cos(π / 4) = 1 / √(2).

 

answered May 13, 2013 by diane Scholar
0 votes

Derivative

g(x) = cosx at x = π / 4

g(π / 4) = cos(π / 4) = 1 / √(2)

The point of the function = (π / 4 , 1 / √(2))

g(x) = cosx

Derivative with respective x

g '(x) = -sinx at  π / 4

g ' (π / 4) = -sin(π / 4) = -1 / √(2).

 

answered May 13, 2013 by diane Scholar

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