How do you find the derivative of  f(x) = 2e^x - 3x^4?

Mar 2, 2018

${f}^{'} \left(x\right) \setminus = \setminus 2 {e}^{x} - 12 {x}^{3}$



Explanation:


The derivative of the given function is represented as:

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) \setminus = \setminus \frac{d}{\mathrm{dx}} \left(2 {e}^{x} - 3 {x}^{4}\right)$



Remember the sum/difference of derivative rule, which is applied when two function are being added/subtracted up.

${\left(f \setminus \pm g\right)}^{'} = {f}^{'} \setminus \pm {g}^{'}$



$= \setminus \frac{d}{\mathrm{dx}} \left(2 {e}^{x}\right) - \setminus \frac{d}{\mathrm{dx}} \left(3 {x}^{4}\right)$

Pull out the constant from the derivative. The rule is stated as:

${\left(a \setminus \cdot f\right)}^{'} = a \setminus \cdot {f}^{'}$

$= 2 \setminus \frac{d}{\mathrm{dx}} \left({e}^{x}\right) - 3 \setminus \frac{d}{\mathrm{dx}} \left({x}^{4}\right)$



Now at this point, we need to recall the power rule for derivative and exponential function rule. They are stated as:

$\setminus \frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$$\text{ }$and$\text{ }$$\setminus \frac{d}{\mathrm{dx}} \left({x}^{a}\right) = a \setminus \cdot {x}^{a - 1}$



So that by applying these rules, we get:

$= 2 {e}^{x} - 3 \setminus \cdot 4 {x}^{4 - 1}$

Simplify to get:

${f}^{'} \left(x\right) \setminus = \setminus 2 {e}^{x} - 12 {x}^{3}$



That's it!