find derivative of
$$f(x)=\frac{ 4 }{ \sqrt{x} }-\frac{ 3 }{ x } + \frac{ 9 }{ x^4 }$$

Question

find derivative of
$$f(x)=\frac{ 4 }{ \sqrt{x} }-\frac{ 3 }{ x } + \frac{ 9 }{ x^4 }$$

chaise · Answer

Remembering that: 

$$\frac{1}{a^{b}}=a^{-b}$$

$$f(x)=4x^{-\frac{1}{2}}-3x ^{-1}+9x ^{-4}$$

You can now use the power rule. If you are still having trouble I might be able to help.

anonymous · Answer

so would it be?
$$-\frac{ 3 }{ 2x^2 }+\frac{ 2 }{ 3x }-\frac{ 5 }{ 36x }$$

anonymous · Answer

actually that can't be the answer in another homework I have multiple choice answers which are

A.$$f'(x)=\frac{ 2 }{ x ^{3/2} }+\frac{ 3 }{ x^2 } \frac{ 36 }{ x^5 }$$
B. $$f'(x)= -\frac{ 2 }{ x ^{3/2} } -\frac{ 3 }{ x^2 } - \frac{ 36 }{ x^3 }$$
C.$$f'(x)=-2\sqrt{x}+\frac{ 3 }{ x^2 }-\frac{ 36 }{ x^3 }$$
D.$$f'(x)=\frac{ 2 }{ x ^{1/2} }-\frac{ 3 }{ x^2 }-\frac{ 36 }{ x^5 }$$

anonymous · Answer

ah i got it
$$f'(x)=-\frac{ 2 }{ x ^{3/2} }+\frac{ 3 }{ x^2 }-\frac{ 36 }{ x^5 }$$