Question

How do you find the derivative of `f(x)=4+x-2x^2` using the limit definition?

Answer 1

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:

1. 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!