d/dx(sinx)
cosx
The derivative of sin(x) with respect to x (dx) is cos(x), which means:
d/dx(sin(x)) = cos(x)
We can prove this using the limit definition of the derivative:
d/dx(sin(x)) = lim(h->0) [sin(x+h) – sin(x)]/h
Now, let’s use the trigonometric identity sin(a+b) = sin(a)cos(b) + cos(a)sin(b) to rewrite sin(x+h):
sin(x+h) = sin(x)cos(h) + cos(x)sin(h)
Substituting this back into the limit definition of the derivative and simplifying we get:
d/dx(sin(x)) = lim(h->0) [(sin(x)cos(h) + cos(x)sin(h)) – sin(x)]/h
= lim(h->0) [sin(x)cos(h)/h + cos(x)sin(h)/h]
= sin(x)lim(h->0) (cos(h)/h) + cos(x)lim(h->0) (sin(h)/h)
Using L’Hopital’s rule we can evaluate these limits:
lim(h->0) (cos(h)/h) = 0
lim(h->0) (sin(h)/h) = 1
Therefore:
d/dx(sin(x)) = cos(x)
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