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anonymous
 4 years ago
can someone please explain to me how the derivative of f(x)=sin x is?? and how its found.
anonymous
 4 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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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0lim as h> 0; sin(x+h)sin(x)  h

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so, theres no cosine involved in the answer.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0sin(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
 4 years ago
Best ResponseYou've already chosen the best response.0\[sin(x)\frac{cos(h)1}{h}+cos(x)\frac{sin(h)}{h}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{cos(h)1}{h} \implies 0;\ and \frac{sin(h)}{h} \implies 1\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.00 + cos(x) = cos(x) Dx(sin(x)) = cos(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[d(\sin x)/dx=\lim_{dx \rightarrow 0} [\sin(x+dx) \sin(x)]/[x+dxx]\] \[d(\sin x)/dx=\lim_{dx \rightarrow 0} [2\cos(x+dx/2)*\sin(dx/2)]/[dx]\] \[= \lim_{dx \rightarrow 0} [\cos(x+dx/2)]*\ [\sin(dx/2)]/[dx/2]\] =cos(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in the change of delta x, why is h by itself , what were the other two variables that were cancelled out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0go for it bro dont be shamed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{a \rightarrow 0} \sin(a)/a=1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i just cant understand it how sin x ends up cos x , the derivative of the function,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sin(A)sin(B)=2*cos((A+B)/2)*sin((AB)/2)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hope you know the general eqn for finding derivatives...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no , LOL, ill get it though thats a promise
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