Why is the derivative of x^n, nx^(n-1)?
From the definition of a derivative: f'(x) = lim h->0 ((f(x+h) - f(x)) / h) Let f(x) = x^n --> d\dx x^n = lim h->0 (((x+h)^n - x^n) / h) By binomial expansion, (x+h)^n = x^n + nhx^(n-1) + n(n-1)h^2 x^(n-2) + ... + h^n --> d\dx x^n = lim h->0 ((x^n + nhx^(n-1) + n(n-1)h^2 x^(n-2) + ... + h^n - x^n) / h) = lim h->0 (nx^(n-1) + n(n-1)h x^(n-2) + ... + h^(n-1)) = nx^(n-1)
JF
Related Maths A Level answers
All answers ▸(i) Find the coordinates of the stationary point on the curve y = 3x^2 − 6/x − 2. [5] (ii) Determine whether the stationary point is a maximum point or a minimum point.
Express (3+ i)(1 + 2i) as a complex number in the form a+bi where a and b are real numbers.
Find the stationary points on the curve: y = x^3 + 3x^2 +2x+5
Determine whether the line with equation 2x+ 3y + 4 = 0 is parallel to the line through the points with coordinates (9, 4) and (3, 8).