Given:
f(x) = cos x
By using the derivative formula,
\( \lim\limits_{h \to 0} \dfrac{cosh-1}{h} \)
\( \lim\limits_{h \to 0} \dfrac{-(1-cosh)}{h} \)
\(- \lim\limits_{h \to 0} \dfrac{2sin^2\dfrac{h}{2}}{h} \)
By using algebra of limits we get
\( - \lim\limits_{h \to 0} \dfrac{sin\dfrac{h}{2}}{\dfrac{h}{2}}\times \lim\limits_{h \to 0} h \)
\( f'(0) = -1\times 0=0\)
Derivative off(x) = cosx at x = 0 is 0
Answered by Aaryan | 1 year ago