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MITL3ARN3R

  • 3 years ago

can someone please explain to me how the derivative of f(x)=sin x is?? and how its found.

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  1. amistre64
    • 3 years ago
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    lim as h-> 0; sin(x+h)-sin(x) ------------ h

  2. MITL3ARN3R
    • 3 years ago
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    so, theres no cosine involved in the answer.

  3. amistre64
    • 3 years ago
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    sin(x+h) = sin(x)cos(h) + sin(h)cos(x) sin(x)cos(h) + sin(h)cos(x) - sin(x) = sin(x)(cos(h)-1) + sin(h)cos(x)

  4. amistre64
    • 3 years ago
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    \[sin(x)\frac{cos(h)-1}{h}+cos(x)\frac{sin(h)}{h}\]

  5. amistre64
    • 3 years ago
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    \[\frac{cos(h)-1}{h} \implies 0;\ and \frac{sin(h)}{h} \implies 1\]

  6. amistre64
    • 3 years ago
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    0 + cos(x) = cos(x) Dx(sin(x)) = cos(x)

  7. Annand
    • 3 years ago
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    \[d(\sin x)/dx=\lim_{dx \rightarrow 0} [\sin(x+dx)- \sin(x)]/[x+dx-x]\] \[d(\sin x)/dx=\lim_{dx \rightarrow 0} [2\cos(x+dx/2)*\sin(dx/2)]/[dx]\] \[= \lim_{dx \rightarrow 0} [\cos(x+dx/2)]*\ [\sin(dx/2)]/[dx/2]\] =cos(x)

  8. MITL3ARN3R
    • 3 years ago
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    in the change of delta x, why is h by itself , what were the other two variables that were cancelled out

  9. MITL3ARN3R
    • 3 years ago
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    go for it bro dont be shamed

  10. amistre64
    • 3 years ago
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    x+h-x .... = h

  11. Annand
    • 3 years ago
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    \[\lim_{a \rightarrow 0} \sin(a)/a=1\]

  12. MITL3ARN3R
    • 3 years ago
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    i just cant understand it how sin x ends up cos x , the derivative of the function,

  13. Annand
    • 3 years ago
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    sin(A)-sin(B)=2*cos((A+B)/2)*sin((A-B)/2)

  14. Annand
    • 3 years ago
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    hope you know the general eqn for finding derivatives...

  15. MITL3ARN3R
    • 3 years ago
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    yeah

  16. Annand
    • 3 years ago
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    got it?

  17. MITL3ARN3R
    • 3 years ago
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    no , LOL, ill get it though thats a promise

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