Can any one canf ind the derivatives of the following functions with all the necessary steps? - canf find external drive
1) Let f (x) = 4 x sin3x derivative f (x) = ______
2) Let g (x) = cos derive (x) = ________
xsquare
3) Let g (x) = g XLNX then II (x) __________
4) Let h (x) = cos x-sin x derivatives, then h (x) = ______
5) Let f (x) = x4 2 xqube-8x 1 then fderivative (x) _________
6) Let f (x) = log x, then derivatives f (x) = _______
7) The derivation of Xsquare 1 in the root
8) The derivative of sin xsquare 1
Wednesday, February 3, 2010
Canf Find External Drive Can Any One Canf Ind The Derivatives Of The Following Functions With All The Necessary Steps?
3 comments:
Note:
"Derivative f (x) is the same as f (x) and the second derivative is f (x)
_______________________________
1)
f (x) = 3x + sin 4x
Since you are = cos u du, and the derivative of 4x 4, then
f '(x) = cos 3x d / dx (3x) + 4
Since the derivative of 3x is 3,
f '(x) = 3 cos 3x + 4
_______________________________
2)
g (x) = cos x
Since d = sin cos u - u, then
g '(x) = - sin x
_______________________________
3)
g (x) = x ln x
We have d ln u = 1 / ud UV = v + u du dv
g '(x) = x ln D / DX (x) + xd / dx (ln x)
Then
g '(x) = ln x (1) + x (1 / x)
Then
g '(x) = ln x + 1
The second derivative is
g (x) = 1 / x
_______________________________
4)
h (x) = cos x - sin x
= [(X ² + 1) d / dx sin (x) - sin XD / dx (x ² + 1)] / (x ² + 1) ²
Then
= Cos [x (x ² + 1) - x sin (2x)] / (x ² + 1) ²
Syndicate
= (X cos ² x + cos x - 2 x sin x) / (x ² + 1) ²
_______________________________
^ _ ^
Note:
"Derivative f (x) is the same as f (x) and the second derivative is f (x)
_______________________________
1)
f (x) = 3x + sin 4x
Since you are = cos u du, and the derivative of 4x 4, then
f '(x) = cos 3x d / dx (3x) + 4
Since the derivative of 3x is 3,
f '(x) = 3 cos 3x + 4
_______________________________
2)
g (x) = cos x
Since d = sin cos u - u, then
g '(x) = - sin x
_______________________________
3)
g (x) = x ln x
We have d ln u = 1 / ud UV = v + u du dv
g '(x) = x ln D / DX (x) + xd / dx (ln x)
Then
g '(x) = ln x (1) + x (1 / x)
Then
g '(x) = ln x + 1
The second derivative is
g (x) = 1 / x
_______________________________
4)
h (x) = cos x - sin x
= [(X ² + 1) d / dx sin (x) - sin XD / dx (x ² + 1)] / (x ² + 1) ²
Then
= Cos [x (x ² + 1) - x sin (2x)] / (x ² + 1) ²
Syndicate
= (X cos ² x + cos x - 2 x sin x) / (x ² + 1) ²
_______________________________
^ _ ^
@ (Sin (3x) 4 x) '= 3cos3x 4
@ Cos (x) '=- sin (x)
@ (XLNX) '= 1 * ln (x) + x * 1 / x = ln (x) +1
@ XLN (())''=( x ln (x) +1) = 1 / x
@ (Cos (x)-sin (x ))'=- sin (x)-cos (x)
@ (X4 +2 x3-8x +1) '= 4x ^ 3 +6 x ^ 2.8
@ (Log (x ))'=( 1/ln10 * ln (x)) '=
(1/ln10) * (1 / x)
@ ((X 2 +1) ^ 1 / 2) '= x / ((x 2 +1) ^ 1 / 2)
@ (Sin (x) / (x2 1 ))'=( cos (x) * (x2 +1)
2xsin ())/(( x2-x +1) ^ 2)
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