# How do you use the definition of a derivative to find the derivative of f(x)=3x^4+2x^3-3x-2?

Feb 3, 2017

$\frac{\mathrm{df}}{\mathrm{dx}} = 12 {x}^{3} + 6 {x}^{2} - 3$

#### Explanation:

The definition of derivative gives $\frac{\mathrm{df}}{\mathrm{dx}} = L {t}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

As $f \left(x\right) = 3 {x}^{4} + 2 {x}^{3} - 3 x - 2$,

f(x+h))=3(x+h)^4+2(x+h)^3-3(x+h)-2

= $3 \left({x}^{4} + 4 {x}^{3} h + 6 {x}^{2} {h}^{2} + 4 x {h}^{3} + {h}^{4}\right) + 2 \left({x}^{3} + 3 {x}^{2} h + 2 x {h}^{2} + {h}^{3}\right) - 3 x - 3 h - 2$

= $3 {x}^{4} + 2 {x}^{3} - 3 x - 2 + h \left(12 {x}^{3} + 6 {x}^{2} - 3\right) + {h}^{2} \left(18 {x}^{2} + 4 x\right) + 12 x {h}^{3} + 3 {h}^{4}$

and $f \left(x + h\right) - f \left(x\right) = h \left(12 {x}^{3} + 6 {x}^{2} - 3\right) + {h}^{2} \left(18 {x}^{2} + 4 x\right) + 12 x {h}^{3} + 3 {h}^{4}$

and $\frac{\mathrm{df}}{\mathrm{dx}} = L {t}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

= $L {t}_{h \to 0} \frac{1}{h} \left(h \left(12 {x}^{3} + 6 {x}^{2} - 3\right) + {h}^{2} \left(18 {x}^{2} + 4 x\right) + 12 x {h}^{3} + 3 {h}^{4}\right)$

= $L {t}_{h \to 0} 12 {x}^{3} + 6 {x}^{2} - 3 + h \left(18 {x}^{2} + 4 x\right) + 12 x {h}^{2} + 3 {h}^{3}$

= $12 {x}^{3} + 6 {x}^{2} - 3$