Power Rule is a corallary to Chain Rule. Power Rule If f(x) = x n then f ' (x) = n x (n-1) Replacing x by g(x) gives.

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Presentation transcript:

Power Rule is a corallary to Chain Rule

Power Rule If f(x) = x n then f ' (x) = n x (n-1) Replacing x by g(x) gives

Power Rule k(x) = g n (x) = [g(x)] n k’(x) = n [g (x)] n-1 g’(x)

Power Rule k(x) = [x 2 + x ] 3 k’(x) = 3 [x 2 + x] 2 (2x+1)

Power Rule k(x) = 2[3x 3 + x ] 4 k’(x) = 8 [3x 3 + x] 3 (9x 2 +1)

k(x) = (x 3 + 2x) 4 k’(x)= A. 4 (x 3 + 2x) 3 (3x 2 + 2) B. 4 (x 3 + 2x) 3 (3x + 2) C. 4 (x 3 + 2x) 3 (3x + 2x)

k(x) = (x 3 + 2x) 4 k’(x)= A. 4 (x 3 + 2x) 3 (3x 2 + 2) B. 4 (x 3 + 2x) 3 (3x + 2) C. 4 (x 3 + 2x) 3 (3x + 2x)

Power Rule k(x) = 2[3x 3 - x -2 ] 20 k’(x) = 40 [3x 3 - x -2 ] 19 (9x 2 +2x -3 )

t(x) = (2x 5 + 3x 2 + 2) 4 t’(x)= A. 4 (2x 5 + 3x 2 + 2) 3 (10x + 6x + 2) B. 4 (10x 4 + 6x) 3 C. 4 (2x 5 + 3x 2 + 2) 3 (10x 4 + 6x)

t(x) = (2x 5 + 3x 2 + 2) 4 t’(x)= A. 4 (2x 5 + 3x 2 + 2) 3 (10x + 6x + 2) B. 4 (10x 4 + 6x) 3 C. 4 (2x 5 + 3x 2 + 2) 3 (10x 4 + 6x)

s(t)=3(t-2/t) 7 = 3(t-2t -1 ) 7 s’(t) = A. 21t - 42 B. 21(t - 2/t) 6 (1 + 2t 0 ) C. 21(t - 2/t) 6 (1 - 2t -1 ) D. 21(t - 2/t) 6 (1 + 2t -2 )

s(t)=3(t-2/t) 7 = 3(t-2t -1 ) 7 s’(t) = A. 21t - 42 B. 21(t - 2/t) 6 (1 + 2t 0 ) C. 21(t - 2/t) 6 (1 - 2t -1 ) D. 21(t - 2/t) 6 (1 + 2t -2 )

Power Rule =[3x 3 - x 2 ] 1/2 =[3x 3 - x 2 ] 1/2 k’(x) = ½ [3x 3 - x 2 ] -1/2 (9x 2 -2x)

rewrite y using algebra. rewrite y using algebra. A.. B.. C..

rewrite y using algebra. rewrite y using algebra. A.. B.. C..

= dy/dx= = dy/dx= A.. B.. C..

= dy/dx= = dy/dx= A.. B.. C..

Power Rule k(x) = [sin x] 13 k’(x) = 13 [sin x] 12 (cos x)

Power Rule k(x) = [csc x] 3 = csc 3 x k’(x) = 3 [csc x] 2 (-csc x cot x)

Power Rule k(x) = [csc x] 3 = csc 3 x k’(x) = 3 [csc x] 2 (-csc x cot x)

Power Rule k(x) = [sin x] -3 = csc 3 x k’(x) = -3 [sin x] -4 (cos x)

t(x) = (tan x) 4 = tan 4 x t’(x)= A. 4 (tan x) 3 B. (tan x) 3 (sec x) 2 C. 4 (tan x) 3 (sec x) 2

t(x) = (tan x) 4 = tan 4 x t’(x)= A. 4 (tan x) 3 B. (tan x) 3 (sec x) 2 C. 4 (tan x) 3 (sec x) 2

..

. A.. B.. C..

. A.. B.. C..

k(x) = ( x 2 + x) 3 Evaluate k’(1) A. 22 B. 24 C. 36 D. 38

t(x) = t’(x)= A.. B.. C. C.

The composition function k(x) = (f o g)(x) = f (g(x)) gR--->[-¼,+oo) f ---->[-1/64, +oo)

Theorem 1 The chain rule If k(x) =f (g(x)), then k’(x) = f ’ ( g(x) ) g’(x)

y = sin(x/2) y = sin(x/2) y = sin (x) y = sin (x)

y = sin(x/4) y = sin(x/4) y = sin (x/2) y = sin (x/2)

y = sin (x) y’ = cos (x) y = sin (x) y’ = cos (x) y = sin (2x) y = sin (2x) y’ = cos (2x) 2 y’ = cos (2x) 2

y = sin(2x) y = sin(4x) y = sin(2x) y = sin(4x) y = sin(4x) y = sin(4x) y’ = cos(4x) 4 y’ = cos(4x) 4

y = tan(4x) y’ = sec 2 (4x) y’ = sec 2 (4x) 4 = 4[sec(4x)] 2.

y = [sin(4x)] 21. y = [sin(4x)] 21. y’ = y’ =

y = [sin(4x)] 21. y = [sin(4x)] 21. y’ = 21[sin(4x)] 20 y’ = 21[sin(4x)] 20

y = [sin(4x)] 21. y = [sin(4x)] 21. y’ = 21[sin(4x)] 20 y’ = 21[sin(4x)]

y = [sin(4x)] 21. y = [sin(4x)] 21. y’ = 21[sin(4x)] 20 cos(4x) y’ = 21[sin(4x)] 20 cos(4x)

y = [sin(4x)] 21. y = [sin(4x)] 21. y’ = 21[sin(4x)] 20 cos(4x) y’ = 21[sin(4x)] 20 cos(4x) Double doink-doink Double doink-doink

y = [sin(4x)] 21. y = [sin(4x)] 21. y’ = 21[sin(4x)] 20 cos(4x)4 y’ = 21[sin(4x)] 20 cos(4x)4 = 84 [sin(4x)] 20 cos(4x)

y = cot(3  x) y = cot(3  x) y’ = –csc 2 (3  x) 3  y’ = –csc 2 (3  x) 3  y = cos(5  x) y = cos(5  x) y = -sin(5  x) 5  y = -sin(5  x) 5  y’ = -5  sin(5  x) y’ = -5  sin(5  x)

If f(x) = sin(10x), find f’(0)

If f(x) = cos(12x), find f’(0)

0.00.1

Theorem : Chain Rule [f o g ]' (x) = f '[g(x)]g'(x).

Chain Rule [f o g ]' (x) = f '[g(x)]g'(x) The derivative of the composite is the derivative of the outside evaluated at the inside times the derivative of the inside evaluated at x

Theorem : Chain Rule [f o g ]' (x) = f '[g(x)]g'(x) [f o g ]' (x) = f '[g(x)]g'(x) Differentiate f(x), replacing x by g(x), differentiate g(x), and multiply.

Theorem : Chain Rule Thus to find y’ y = sin(x 2 ) y' = cos(x 2 ) (2x)

y(sqrt(  )/2) = sin(  ) = sqrt(2)/2 y'(sqrt(  )/2) = cos(  /4) 2 sqrt(  )/2 = sqrt(2)/2 [2] sqrt(  )/2 = sqrt(2  )/2 y = sin(x 2 ) y’ = cos(x 2 )2x

Theorem : Chain Rule y' = cos(x 2 ) (2x) and when x = sqrt(  )/2, y'(sqrt(  )/2) = cos(  /4) 2 sqrt(  )/2 = sqrt(2)/2 [2] sqrt(  )/2 = sqrt(2  )/2 The equation of the tangent line would be The equation of the tangent line would be [y - sqrt(2)/2] / [x - sqrt(  )/2] = sqrt(2  )/2 [y - sqrt(2)/2] / [x - sqrt(  )/2] = sqrt(2  )/2

Differentiate f(x), replacing x by g(x), differentiate g(x), and differentiate g(x), and multiply the preceeding answers together. multiply the preceeding answers together.

If f(x) = x n then [f(g(x))]’= f ' (g(x))g’(x) = n g(x) n-1 g’(x) y = (csc(x)+x 3 ) 5 then y’ = 5(csc(x)+x 3 ) 4

If f(x) = x n then [f(g(x))]’= f ' (g(x))g’(x) = n g(x) n-1 g’(x) y = (csc(x)+x 3 ) 5 then y’ = 5(csc(x)+x 3 ) 4 (-csc(x)cot(x)+3x 2 ) and y’ (  /2) =

If f(x) = x n then [f(g(x))]’= f ' (g(x))g’(x) = n g(x) n-1 g’(x) y = (csc(x)+x 3 ) 5 then y’ = 5(csc(x)+x 3 ) 4 (-csc(x)cot(x)+3x 2 ) and y’(  /2) = 5(1 + (  /2) 3 ) 4 (0 + 3(  /2) 2 )