Asked by Riley
I need help simplifying this problem.
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
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
Answers
Answered by
Reiny
your 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]
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]
Answered by
kartik
prove that sin invers (1/5)+COT INVERSE (3)=0
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