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March 1, 2015

March 1, 2015

Posted by **Riley** on Tuesday, March 31, 2009 at 10:52am.

f(x)=x^4/5(x-4)^2

f'(x)=x^4/5*2(x-4)+(x-4)^2*4/5x^-1/5

it is suposed to end up as this:

1/5x^-1/5(x-4)[5*x*2+(x-4)*4]

but how do i get it to there? I need to see all the steps and how to get them. Thanks

- calculus -
**Reiny**, Tuesday, March 31, 2009 at 12:12pmyour first line is good

now, do you notice x^(-1/5) and x^(4/5) ?

isn't x^(-1.5) the highest common factor ? (just like a^4 would be the HCF for a^4 and a^6)

also (x-4) is a common factor,

as well as -1/5 (pretend your first term was (5/5)x^4/5*2(x-4)

so

(-1/5)(x^(-1/5)(x-4)[10 + 4(x-4)]

= (-1/5)(x^(-1/5)(x-4)[4x-6]

= (-2/5)(x^(-1/5)(x-4)[2x-3]

- calculus -
**kartik**, Sunday, August 1, 2010 at 11:50amprove that sin invers (1/5)+COT INVERSE (3)=0

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