Plss i need solution or maybe an answer i just want to make sure if my solutions are right here the question its Calculus f(x)''''' of f(x)=15x^4-(2x-1/2)^6

Question

Plss i need solution or maybe an answer i just want to make sure if my solutions are right here the question its Calculus >>>f(x)''''' of f(x)=15x^4-(2x-1/2)^6

misty1212 · Answer

lol hi!

misty1212 · Answer

is that the fifth derivative?

anonymous · Answer

$$ f(x)=15x ^{4}-(\frac{ 2x-1 }{ 2 })^{6}$$

anonymous · Answer

Yup

anonymous · Answer

i need some solutions or answers already :)

kainui · Answer

Do you know how to take the first derivative?

anonymous · Answer

Yup i already got the solutions :) i just want to compare if my answers correct

anonymous · Answer

the degree of the first term is 4 so it will disappear in the fifth derivative.

anonymous · Answer

Yeah that my answer is it right?

anonymous · Answer

-360(2x-1)

anonymous · Answer

the second terms will leave this after 5 derivatives

anonymous · Answer

then its -720x+360 right ?

anonymous · Answer

correct :)

anonymous · Answer

the sixth derivative leaves -720 and seventh derivative leaves 0

anonymous · Answer

That professor !!!! -_- she said im wrong :(

anonymous · Answer

Our teacher said that my answers wrong?

anonymous · Answer

$$f'(x) =15\times 4\times x^3 - 6(x-\frac{1}{2})^5$$
$$f''(x) =15\times4\times 3 x^2 - 6\times5(x-\frac{1}{2})^4$$

anonymous · Answer

Umm.. Its ok i know already how to get it i just want know if im wrong .. ^^ Thanks

anonymous · Answer

$$f'''(x) =15\times4\times3 \times2x - 6\times5\times4(x-\frac{1}{2})^3$$
$$f''''(x) =15\times4\times3 \times2\times1 - 6\times5\times4\times3(x-\frac{1}{2})^2$$$$f'''''(x) =15\times4\times3 \times2\times1\times0 - 6\times5\times4\times3\times2(x-\frac{1}{2})^1$$ =0 - 720x+360
Your answer is absolutely correct. (according to leibnitz)

anonymous · Answer

:)