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

Mathematics
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lim as h-> 0; sin(x+h)-sin(x) ------------ h
so, theres no cosine involved in the answer.
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)

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Other answers:

\[sin(x)\frac{cos(h)-1}{h}+cos(x)\frac{sin(h)}{h}\]
\[\frac{cos(h)-1}{h} \implies 0;\ and \frac{sin(h)}{h} \implies 1\]
0 + cos(x) = cos(x) Dx(sin(x)) = cos(x)
\[d(\sin x)/dx=\lim_{dx \rightarrow 0} [\sin(x+dx)- \sin(x)]/[x+dx-x]\] \[d(\sin x)/dx=\lim_{dx \rightarrow 0} [2\cos(x+dx/2)*\sin(dx/2)]/[dx]\] \[= \lim_{dx \rightarrow 0} [\cos(x+dx/2)]*\ [\sin(dx/2)]/[dx/2]\] =cos(x)
in the change of delta x, why is h by itself , what were the other two variables that were cancelled out
go for it bro dont be shamed
x+h-x .... = h
\[\lim_{a \rightarrow 0} \sin(a)/a=1\]
i just cant understand it how sin x ends up cos x , the derivative of the function,
sin(A)-sin(B)=2*cos((A+B)/2)*sin((A-B)/2)
hope you know the general eqn for finding derivatives...
yeah
got it?
no , LOL, ill get it though thats a promise

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