Question

How do you use the definition of a derivative to find the derivative of `f(x)=(x^2+3)(2x-7)`?

Answer 1

`dy/dx=6x^2-14x+6`

Explanation:

Given -

`f(x) = (x^2+3)(2x-7)`

We have to apply product rule.

`u=(x^2+3)`
`v=(2x-7)`

`dy/dx =u.(dv)/dx+v.(du)/dx`

`dy/dx=[(x^2+3)(2)]+[(2x-7)(2x)]`

`dy/dx=[2x^2+6]+[4x^2-14x]`

`dy/dx=2x^2+6+4x^2-14x`

`dy/dx=6x^2-14x+6`

Answer 2

I would recommend that you multiply the function out and then use the formula.

`f(x) = 2x^3 - 7x^2 + 6x - 21`

Now apply the formula `f'(x) = lim_(h->0) (f(x + h) - f(x))/h`.

`f'(x) = lim_(h->0) (2(x + h)^3 - 7(x + h)^2 + 6(x +h) - 21 - (2x^3 - 7x^2 + 6x - 21))/h`

`f'(x) = lim_(h->0) (2(x + h)(x + h)(x + h) - 7(x^2+ 2xh + h^2) + 6x + 6h - 21 - 2x^3 + 7x^2 - 6x + 21))/h`

`f'(x) = lim_(h->0) (2(x^2 + 2xh + h^2)(x + h) - 7x^2 - 14xh - 7h^2 + 6x + 6h - 21 - 2x^3 +7x^2 - 6x + 21)/h`

`f'(x) = lim_(h->0) (2(x^3 + 2x^2h + h^2x + hx^2 + 2xh^2 + h^3) - 7x^2 - 14xh - 7h^2 + 6x + 6h - 21 - 2x^3 + 7x^2 - 6x + 21)/h`

`f'(x) = lim_(h->0) (2x^3 + 4x^2h + 2h^2x + 2hx^2 + 4xh^2 + 2h^3 - 7x^2 - 14xh - 7h^2 + 6x+ 6h - 21 - 2x^3 + 7x^2 - 6x +21)/h`

Now notice what cancels out.

`f'(x) = lim_(h->0) (4x^2h + 2h^2x + 2hx^2 + 4xh^2 + 2h^3 - 14xh - 7h^2 + 6h)/h`

Factor out an `h`:

`f'(x) = lim_(h->0) (h(4x^2 + 2hx + 2x^2 + 2xh + 2h^2 - 14x - 7h + 6))/h`

`f'(x) = lim_(h->0) 6x^2 + 4xh + 2h^2 - 14x - 7h + 6`

`f'(x) = 6x^2 + 4x(0) + 2(0)^2 - 14x - 7(0) + 6`

`f'(x) = 6x^2 - 14x + 6`

If we verify using the power rule that this is correct we get:

`d/dx(2x^3 -7x^2 + 6x -21) = 2(3)x^2 - 7(2)x + 6(1) = 6x^2 - 14x + 6`

Which is what we got using the definition of the derivative.

Hopefully this helps!