For problems like this, use the formula #ƒ'(x) = lim_(h -> 0) (ƒ(x + h) - ƒ(x))/h#.
#f'(x) = lim_(h -> 0) (-2(x + h)^2 + 1(x + h) + 4 - (-2x^2 + x + 4))/h#
#f'(x) = lim_(h -> 0) (-2(x^2 + 2xh + h^2) + x + h + 4 + 2x^2 - x - 4)/h#
#f'(x) = lim_(h -> 0) (-2x^2 - 4xh - 2h^2 + x + h+ 4 + 2x^2 - x - 4)/h#
#f'(x) = lim_(h-> 0) (-4xh - 2h^2 + h)/h#
#f'(x) = lim_(h -> 0) (cancel(h)(-4x - 2h + 1))/(cancel(h))#
#f'(x) = -4x - 2(0) + 1#
#f'(x) = -4x + 1#
The derivative of #f(x) = 4 + x - 2x^2# is #f'(x) = -4x + 1#, or #dx/dy = -4x + 1# in alternate notation.
Practice exercises:
- Determine the derivatives of the following functions.
a) #f(x) = -3x^2 + 7#
b) #g(x) = 1/2x^2 + 6x - 4#
#2.# The formula #f'(a) = lim_(h ->0)(f(a + h) - f(a))/h#, where a is a point on the graph of a function, is used to determine the slope of the line tangent at that point. Determine the slopes of the tangents of the following functions, ant the given point.
a) #f(x) = 3x^2 - 4x, x = -1#
b) #g(x) = -1/(2x), x = 3#
Hopefully this helps, and good luck!
