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-   -   2nd, 3rd and 4th derivatives (http://mymathforum.com/calculus/25975-2nd-3rd-4th-derivatives.html)

 cacophonyjm March 20th, 2012 10:47 PM

2nd, 3rd and 4th derivatives

My attempt at calculating these:

a) d^2/(dx^2): ln(x+1)
b) d^3/(dx^3): x^7 + 4x^6 - x^2
c) d^2/(dx^2): 1/(x + 1)
d) d^4/(dx^4): cos(2x)

a) Product rule: 1/x(x + 1) + ln(1)= 1 + 1/x + ln
Linearity rule: -(1/x^2) + 1/x

b) Linearity rule: 7x^6 + 6x^4 - 2x= 6(7x)^5 + 4(6x)^3 - 2=
(5)(6)(7x)^4 + (3)(4)(6x)^2 = 210x^4 + 72x^2

c) Quotient rule: ((0)(x + 1) - (1)(1))/(x + 1)^2=
-1/(x + 1)^2=
((0)(x+1)^2 - (-1)2(x + 1))/[(x + 1)^2]^2=
(-2x - 2)/(x + 1)^4

d) the answer ends up being: cos(2x) + sin(2) I think

Do these look valid?

 MarkFL March 20th, 2012 11:13 PM

Re: 2nd, 3rd and 4th derivatives

a) $\frac{d^2}{dx^2}$$\ln(x+1)$$=\frac{d}{dx}$$\frac{d }{dx}\(\ln(x+1)$$\)=\frac{d}{dx}$$\frac{1}{x+1}$$=-\frac{1}{(x+1)^2}$

b) $\frac{d^3}{dx^3}$$x^7+4x^6-x^2$$=\frac{d}{dx}$$\frac{d}{dx}\(\frac{d}{dx}\(x^ 7+4x^6-x^2$$\)\)=$

$\frac{d}{dx}$$\frac{d}{dx}\(7x^6+24x^5-2x$$\)=\frac{d}{dx}$$42x^5+120x^4-2$$=$

$210x^4+480x^3=30x^3$$7x+16$$$

c) $\frac{d^2}{dx^2}$$\frac{1}{x+1}$$=\frac{d}{dx}$$\f rac{d}{dx}\(\frac{1}{x+1}$$\)=\frac{d}{dx}$$-\frac{1}{(x+1)^2}$$=\frac{2}{(x+1)^3}$

d) $\frac{d^4}{dx^4}$$\cos(2x)$$=2^4\cos(2x)=16\cos(2x )$

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