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

Apr 8, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} - 14 x + 6$

#### Explanation:

Given -

$f \left(x\right) = \left({x}^{2} + 3\right) \left(2 x - 7\right)$

We have to apply product rule.

$u = \left({x}^{2} + 3\right)$
$v = \left(2 x - 7\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = u . \frac{\mathrm{dv}}{\mathrm{dx}} + v . \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\left({x}^{2} + 3\right) \left(2\right)\right] + \left[\left(2 x - 7\right) \left(2 x\right)\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[2 {x}^{2} + 6\right] + \left[4 {x}^{2} - 14 x\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{2} + 6 + 4 {x}^{2} - 14 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} - 14 x + 6$

Apr 8, 2017

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

$f \left(x\right) = 2 {x}^{3} - 7 {x}^{2} + 6 x - 21$

Now apply the formula $f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$.

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{2 {\left(x + h\right)}^{3} - 7 {\left(x + h\right)}^{2} + 6 \left(x + h\right) - 21 - \left(2 {x}^{3} - 7 {x}^{2} + 6 x - 21\right)}{h}$

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

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{2 \left({x}^{2} + 2 x h + {h}^{2}\right) \left(x + h\right) - 7 {x}^{2} - 14 x h - 7 {h}^{2} + 6 x + 6 h - 21 - 2 {x}^{3} + 7 {x}^{2} - 6 x + 21}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{2 \left({x}^{3} + 2 {x}^{2} h + {h}^{2} x + h {x}^{2} + 2 x {h}^{2} + {h}^{3}\right) - 7 {x}^{2} - 14 x h - 7 {h}^{2} + 6 x + 6 h - 21 - 2 {x}^{3} + 7 {x}^{2} - 6 x + 21}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{2 {x}^{3} + 4 {x}^{2} h + 2 {h}^{2} x + 2 h {x}^{2} + 4 x {h}^{2} + 2 {h}^{3} - 7 {x}^{2} - 14 x h - 7 {h}^{2} + 6 x + 6 h - 21 - 2 {x}^{3} + 7 {x}^{2} - 6 x + 21}{h}$

Now notice what cancels out.

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{4 {x}^{2} h + 2 {h}^{2} x + 2 h {x}^{2} + 4 x {h}^{2} + 2 {h}^{3} - 14 x h - 7 {h}^{2} + 6 h}{h}$

Factor out an $h$:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{h \left(4 {x}^{2} + 2 h x + 2 {x}^{2} + 2 x h + 2 {h}^{2} - 14 x - 7 h + 6\right)}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} 6 {x}^{2} + 4 x h + 2 {h}^{2} - 14 x - 7 h + 6$

$f ' \left(x\right) = 6 {x}^{2} + 4 x \left(0\right) + 2 {\left(0\right)}^{2} - 14 x - 7 \left(0\right) + 6$

$f ' \left(x\right) = 6 {x}^{2} - 14 x + 6$

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

$\frac{d}{\mathrm{dx}} \left(2 {x}^{3} - 7 {x}^{2} + 6 x - 21\right) = 2 \left(3\right) {x}^{2} - 7 \left(2\right) x + 6 \left(1\right) = 6 {x}^{2} - 14 x + 6$

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

Hopefully this helps!