Why Are My Calculus Derivatives Incorrect?

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Homework Statement


f(x)=X^2/(x^2-16)

f(x)=1+x/1-X

f(x)=X^3(X-2)^2

Ive done the first and second derivatives but they just don't seem right

Homework Equations


Quotient/Chain/Product Rule


The Attempt at a Solution


f(x)=X^2/(x^2-16)
(X^2-16)(2X)-(X^2)(2X)/(X^2-16)^2
f'(x)=-32X^2/(X^2-16)^2
(X^4-32X^2+256)(-64X)-(-32X^2)(4X^3-64X)/(X^4-32X^2+256)^2
-64x^5+2048X^3-16384X+128X^5-2048X^3
f''(x)=64X^5-16384X/(X^4-32X^2+256)^2

f(x)=1+x/1-X
(1-X)(1)-(1+X)(-1)/(1-X)^2
f'(x)=2-2X/(1-X)^2
(X^2-2X+1)(-2)-(2X-2)(2-2X)/(X^2-2X+1)^2
f''(x)=2X^2+4X+2/(X^2-2X+1)^2

f(x)=X^3(X-2)^2
(X^3)(2X-4)+(3X^2)(X-2)^2
2X^4-4X^3+3X^4-12X^3+12X^2
f'(x)=5X^4-16X^3+12X^2
5X^4-16^3+12X^2
f''(x)=20X^3-48X^2+24X
 
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hi mmajames! :smile:

eugh! this is impossible to read! :redface:

try using the X2 button just above the Reply box (and a few more spaces) :wink:
 
1. Are these three different functions??

2. Please use parentheses 1+ x/x+ 3 is not the same as (1+ x)/(x+ 3).
 
f(x)=X2/(x2-16)
(X2-16)(2X)-(X2)(2X)/(X2-16)2
f'(x)=-32X2/(X-16)2
(X4-32X4+256)(-64X)-(-32X2)(4X3-64X)/(X4-32X2+256)2
-64x^5+2048X^3-16384X+128X5-2048X3
f''(x)=64X5-16384X/(X4-32X^2+256)2

f(x)=(1+x)/(1-X)
(1-X)(1)-(1+X)(-1)/(1-X)2
f'(x)=2-2X/(1-X)2
(-2X+1)(-2)-(2X-2)(2-2X)/(X2-2X+1)2
f''(x)=2X2+4X+2/(X2-2X+1)22

f(x)=X3(X-2)2
(X3)(2X-4)+(3X2)(X-2)2
2X4-4X3+3X4-12X3+12X2
f'(x)=5X4-16X3+12X2
5X^4-163+12X2
f''(x)=20X3-48X2+24X



And yes Three different functions
 
mmajames said:
f(x)=X2/(x2-16)
(X2-16)(2X)-(X2)(2X)/(X2-16)2
f'(x)=-32X2/(X-16)2

hmm … i can see two mistakes already :redface:

sorry, but I'm not checking the rest until you've gone over them again, to find any other mistakes

also, (as HallsofIvy :smile: says) please use more brackets
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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