Using the limit definition, how do you differentiate f(x)=3x^5 + 4x?

Mar 25, 2016

$f ' \left(x\right) = 15 {x}^{4} + 4$

Explanation:

For theegiven function:
$f \left(x\right) = 3 {x}^{5} + 4 x$

We take the limit definition for a derivative:
$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

This gives:
$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\left(3 {\left(x + h\right)}^{5} + 4 \left(x + h\right)\right) - \left(3 {x}^{5} + 4 x\right)}{h}$

$= {\lim}_{h \rightarrow 0} \frac{3 \left(\cancel{{x}^{5}} + 5 {x}^{4} h + 10 {x}^{3} {h}^{2} + 10 {x}^{2} {h}^{3} + 5 x {h}^{4} + {h}^{5}\right) + \cancel{4 x} - \cancel{3 {x}^{5}} - \cancel{4 x} + 4 h}{h}$

$= {\lim}_{h \rightarrow 0} 3 \left(5 {x}^{4} + 10 {x}^{3} h + 10 {x}^{2} {h}^{2} + 5 x {h}^{3} + {h}^{4}\right) + 4$

Taking the limit as $h$ tends to zero on this expression gives us:
$f ' \left(x\right) = 15 {x}^{4} + 4$