f(x) = sinx. Using differentiation from first principles find the exact value of f' (π/6).

The derivative of the function where x=π/6 is defined asThe limit as h->0 of [sin(h+π/6)-sin(π/6)]/hUsing the double angle formula, sin(h+π/6) = sin(h)cos(π/6) + cos(h)sin(π/6) = √3sin(h)/2 + cos(h)sin(π/6)The limit becomes [sin(h)/2 + cos(h)sin(π/6)-sin(π/6)]/hThe limit can be broken up into two partslim as h->0 of [cos(h)sin(π/6)-sin(π/6)]/h = 0 (could use l'Hospital's rule or half angle formula)lim as h->0 of [√3sin(h)/2]/h = 1/2 (small angle approximation)0+√3/2=√3/2

Related Maths A Level answers

All answers ▸

Integrate lnx


Given that y = (3x^4 + x)^5, find dy/dx using the chain rule.


Differentiate y=ln(2x^2) with respect to x


Find tan(A-B) sec^2(A) - 2tan(A) = 16 && sin(B)sec^2(B) = 64cos(B)cosec^2(B)