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Derivative of cosine

User Mosi
by
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1 Answer

6 votes

Answer:

= -sin(x)

Explanation:

Possible derivation:

d/dx(cos(x))

From the limit definition of the derivative, d/dx(cos(x)) = lim_(h->0) (cos(x + h) - cos(x))/h:

= lim_(h->0) (cos(x + h) - cos(x))/h

Apply the cosine angle addition formula to cos(x + h):

= lim_(h->0) ((cos(x) cos(h) - sin(x) sin(h)) - cos(x))/h

Collect in terms of cos(x) and sin(x):

= lim_(h->0)(cos(x) (cos(h) - 1)/h - sin(x) sin(h)/h)

Multiply numerator and denominator of (cos(h) - 1)/h by the conjugate term cos(h) + 1 and expand the numerator:

= lim_(h->0)(cos(x) (cos^2(h) - 1)/((cos(h) + 1) h) - sin(x) sin(h)/h)

Apply the Pythagorean identity sin^2(h) + cos^2(h) = 1:

= lim_(h->0)(-cos(x) (sin^2(h))/((cos(h) + 1) h) - sin(x) sin(h)/h)

Factor inside the limit:

= lim_(h->0)((-(cos(x) sin(h))/(cos(h) + 1) - sin(x)) sin(h)/h)

The limit of a product is the product of the limits:

= (lim_(h->0)(-(cos(x) sin(h))/(cos(h) + 1) - sin(x))) (lim_(h->0) sin(h)/h)

By continuity, lim_(h->0)(-(cos(x) sin(h))/(cos(h) + 1) - sin(x)) = -(cos(x) sin(0))/(cos(0) + 1) - sin(x) = -sin(x):

= -sin(x) lim_(h->0) sin(h)/h

Apply the common limit lim_(h->0) sin(h)/h = 1:

Answer: = -sin(x)

User Raymond Hettinger
by
7.6k points