**Step 1:** d/dxcos(x) = lim_(h->0) (cos(x+h) -cos(x) )/h

*Converting the derivative to the difference quotient*

**Step 2:** cos(x+h) = cos(x)cos(h) - sin(x)sin(h)

*Using an identity to expand out cos(x+h)*

**Step 3:** lim_(h->0) (cos(x+h) -cos(x) )/h= lim_(h->0) ((cos(x)cos(h) - sin(x)sin(h)) -cos(x) )/h

*replacing cos(x+h) with the expanded form*

**Step 4:** lim_(x->h) ((cos(x)cos(h) - sin(x)sin(h)) -cos(x) )/h =lim_(x->h) (cos(x)cos(h) - cos(x) - sin(x)sin(h) )/h

*Using Commutative Property To Rearrange The Equation*

**Step 5:** = (lim_(x->h) (cos(x)cos(h) - cos(x)) / h) - (lim_(x->h) (sin(x)sin(h))/h)

*Limit of the Sum is equal to the sum of the limits*

**Step 6:** = (cos(x) lim_(x->h) (1*cos(h) - 1) / h) - (lim_(x->h) (sin(x)sin(h))/h)

*Factoring out a cos(x) from the first half of the equation*

**Step 7:** = (cos(x) lim_(x->h) (1*cos(h) - 1) / h) - (sin(x) lim_(x->h) (1*sin(h))/h)

*Factoring out a sin(x) from the second half of the equation*

**Step 8:** = cos(x) * 0 - sin(x) *1

*Using the squeeze theorem, we know for a fact that* lim_(theta->0) (cos(theta) - 1) / theta = 0 *and that* lim_(theta->0) (sin(theta)) / theta = 1 *we can simplify some of our terms*

**Step 9:** cos(x) * 0 - sin(x) *1 = -sin(x)

*sprinkle in some simplification …*

**Step 10:** d/dx cos(x) = -sin(x)

*We can finally determine that the derivative of cos(x) is indeed equal to -sin(x)*

Some help was received to understand how the squeeze theorem applies to the lim_(theta->0) (cos(theta) - 1) / theta