derivative of sinx+cosx?

Question

Determine all significant features by hand and sketch a graph.

f(x)=sinx+cosx

basically, I need to find the first derivative of the function, get x-values, and determine whether increasing or decreasing. Then take a second derivative and determine whether it's concave up/down.

umm...could you help me find the critical points, etc?

Answer 1

f(x) = sin x + cosx
f `(x) = cos x - sin x
f "(x) = - sin x - cos x

f `(x) = 0 for turning point
cos x = sin x
tan x = 1
x = π/4 , 5π/4 for turning points
f(π/4) = sin π/4 + cos π/4 = 2 / √2
f(5π/4) = sin 5π/4 + cos 5π/4 = - 2 / √2
f "(π/4) is - ve so MAX turning point at π/4
f "(5π/4) is +ve so MIN turning point at 5π/4
f(0) = cos 0 = 1

Answer 2

f(x)=sinx+cosx
diff wrt x
f'(x)= cosx-sinx

f'(x)=0
cosx=sinx
sinx/cosx=1
tanx=1
tanx=tan pi/4
x=pi/4

f"(x)= sinx+cosx
f"(pi/4)= sin pi/4+cos pi/4
= √3/2+√3/2
= √3

Answer 3

f'(x)=cosx-sinx
f''(x)= -sinx-cosx

because the derivative of sinx = cosx
and the derivative of cosx= -sinx
by definition.

set the equations =0 to find your critical points, increasing/decreasing, concave up/concave down.

Answer 4

f(x) = sinx +cosx
f'(x) = cos x -sin x
Slope of the curve = f'(x) = cos x -sin x = 0
cos x =sin x
x = pi/4
f''(x) = -sin x - cos x
f''(pi/4) = - sin pi/4 - cos pi/4 = -1/sq rt 2 -1/sq rt 2 = - 2/sq rt 2
f''(pi/4) = -sq rt 2 < 0
Hence f is maximum at x = pi/2
Hence the function is increasing till x= pi/2 and then decreasing. So it is concave up

Answer 5

cosx-sinx

Answer 6

d/dx(sin x)=cosx
d/dx(cos x)= -sin x
so f'(x)=cos x -sin x

Answer 7

y = sin x + cos x
y ' = cos x - sin x

Answer 8

f '(X) -cosx+sinx

Answer 9

first derivative
cos(x) - sin(x)
second derivative
-sin(x) - cos(x)

Answer 10

f(x) = sin x + cos x

f'(x) = cos x - sin x