Question

please help me in calculus
find the derivative of cosx/x=y

Answer 1

quotient rule for this one
\[\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}\] with
\[f(x)=\cos(x), f'(x)=-\sin(x), g(x)=x, g'(x)=1\]

Answer 2

i got a -sinx/x -cosx/x^2

Answer 3

i get
\[\frac{-x\sin(x)-\cos(x)}{x^2}\]

Answer 4

same think you got, i just have it over one denominator not two

Answer 5

can u please explain it to me i have a test tomorrow and i need to know how to do this

Answer 6

you did it right, we got the same answer

Answer 7

\[\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}\]
\[\left(\frac{\cos(x)}{x}\right)'=\frac{x(-\sin(x))-1\times\cos(x)}{x^2}\]

Answer 8

clean it up and get
\[\frac{-x\sin(x)-\cos(x)}{x^2}\]

Answer 9

good luck on your test
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Answer 10

thank u so much :)

Answer 11

yw