## Calculus: Early Transcendentals 8th Edition

The derivative of $f(x) = cosx$ does equal $f'(x) = -sinx$
Using the definition of the derivative: $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$ $f(x) = cos x$ $f'(x) = \lim\limits_{h \to 0} \frac{cos(x+h) - cos(x)}{h}$ $cos(A + B) = cos(A) \times cos(B) - sin (A) \times sin(B)$ $f'(x) = \lim\limits_{h \to 0} \frac{cosx \times cosh - sinx \times sinh - cosx}{h}$ $f'(x) = \lim\limits_{h \to 0} \frac{cosx(cosh -1) - sinx \times sinh}{h}$ $f'(x) = cosx \times[\lim\limits_{h \to 0}\frac{cosh -1}{h}] - sinx \times [\lim\limits_{h \to 0} \frac{sin h}{h}]$ Now using the terms given by the book: $\lim\limits_{x \to 0} = \frac{cosx-1}{x} = 0$ $\lim\limits_{x \to 0} = \frac{sinx}{x} = 1$ $f'(x) = cosx \times[\lim\limits_{h \to 0}\frac{cosh -1}{h}] - sinx \times [\lim\limits_{h \to 0} \frac{sin h}{h}]$ $f'(x) = cosx \times[0] - sinx \times [1]$ $f'(x) = - sinx$