MITL3ARN3R
can someone please explain to me how the derivative of f(x)=sin x is?? and how its found.



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amistre64
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lim as h> 0; sin(x+h)sin(x)

h

MITL3ARN3R
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so, theres no cosine involved in the answer.

amistre64
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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)

amistre64
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\[sin(x)\frac{cos(h)1}{h}+cos(x)\frac{sin(h)}{h}\]

amistre64
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\[\frac{cos(h)1}{h} \implies 0;\ and \frac{sin(h)}{h} \implies 1\]

amistre64
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0 + cos(x) = cos(x)
Dx(sin(x)) = cos(x)

Annand
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\[d(\sin x)/dx=\lim_{dx \rightarrow 0} [\sin(x+dx) \sin(x)]/[x+dxx]\]
\[d(\sin x)/dx=\lim_{dx \rightarrow 0} [2\cos(x+dx/2)*\sin(dx/2)]/[dx]\]
\[= \lim_{dx \rightarrow 0} [\cos(x+dx/2)]*\ [\sin(dx/2)]/[dx/2]\]
=cos(x)

MITL3ARN3R
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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

MITL3ARN3R
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go for it bro dont be shamed

amistre64
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x+hx .... = h

Annand
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\[\lim_{a \rightarrow 0} \sin(a)/a=1\]

MITL3ARN3R
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i just cant understand it how sin x ends up cos x , the derivative of the function,

Annand
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sin(A)sin(B)=2*cos((A+B)/2)*sin((AB)/2)

Annand
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hope you know the general eqn for finding derivatives...

MITL3ARN3R
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yeah

Annand
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got it?

MITL3ARN3R
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no , LOL, ill get it though thats a promise