# How do you differentiate y = 2^xlog_2(x^4)?

Mar 1, 2016

$f = {2}^{x} , g = {\log}_{2} \left({x}^{4}\right) \to f ' = {2}^{x} \ln 2 , g ' = \frac{1}{{x}^{4} \ln 2} \cdot 4 {x}^{3}$
$y ' = f g ' + g f ' = \frac{{2}^{x} 4 {x}^{3}}{{x}^{4} \ln 2} + \left({2}^{x} \ln 2\right) \left({\log}_{2} {x}^{4}\right)$

#### Explanation:

Use the product rule to differentiate y. Call the first one be f and the second one be g and take the derivatives of each of them then put it in to the product rule

Mar 1, 2016

$\setminus \frac{d}{\mathrm{dx}} \left({2}^{x} \setminus {\log}_{2} \left({x}^{4}\right)\right) = \setminus \frac{{2}^{x} \left(x \setminus {\log}_{2} \left({x}^{4}\right) \setminus {\ln}^{2} \left(2\right) + 4\right)}{x \setminus \ln \left(2\right)}$

#### Explanation:

$\setminus \frac{d}{\mathrm{dx}} \left({2}^{x} \setminus {\log}_{2} \left({x}^{4}\right)\right)$

Applying product rule as: ${\left(f \setminus \cdot g\right)}^{'} = {f}^{'} \setminus \cdot g + f \setminus \cdot {g}^{'}$
$f = {2}^{x} , g = \setminus {\log}_{2} \left({x}^{4}\right)$

$= \setminus \frac{d}{\mathrm{dx}} \left({2}^{x}\right) \setminus {\log}_{2} \left({x}^{4}\right) + \setminus \frac{d}{\mathrm{dx}} \left(\setminus {\log}_{2} \left({x}^{4}\right)\right) {2}^{x}$

$\setminus \frac{d}{\mathrm{dx}} \left({2}^{x}\right) = {2}^{x} \setminus \ln \left(2\right)$

$\setminus \frac{d}{\mathrm{dx}} \left(\setminus {\log}_{2} \left({x}^{4}\right)\right) = \setminus \frac{4}{x \setminus \ln \left(2\right)}$

finally,
$= {2}^{x} \setminus \ln \left(2\right) \setminus {\log}_{2} \left({x}^{4}\right) + \setminus \frac{4}{x \setminus \ln \left(2\right)} {2}^{x}$

Simplifying,
$= \setminus \frac{{2}^{x} \left(x \setminus {\log}_{2} \left({x}^{4}\right) \setminus {\ln}^{2} \left(2\right) + 4\right)}{x \setminus \ln \left(2\right)}$

Mar 1, 2016

To solve this problem it would help if we remember these two intermediate results (proof given below),
Result 1: $\setminus \frac{d}{\mathrm{dx}} {a}^{x} = {a}^{x} \setminus \ln \left(a\right)$
Result 2: $\setminus \frac{d}{\mathrm{dx}} \setminus {\log}_{b} \left({x}^{n}\right) = \setminus \frac{n}{x \setminus \ln \left(b\right)}$

y = 2^x\log_2(x^4);

Applying the Chain Rule,
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \left[\setminus \frac{d}{\mathrm{dx}} \left({2}^{x}\right)\right] \setminus {\log}_{2} \left({x}^{4}\right) + {2}^{x} \left[\setminus \frac{d}{\mathrm{dx}} \setminus {\log}_{2} \left({x}^{4}\right)\right]$

Apply the two results mentioned above,
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \left[{2}^{x} \setminus \ln \left(2\right)\right] {\log}_{2} \left({x}^{4}\right) + {2}^{x} \left[\setminus \frac{4}{x \setminus \ln \left(2\right)}\right]$
$\setminus q \quad \setminus \quad = \setminus \ln \left(2\right) y + \setminus \frac{4}{\setminus \ln \left(2\right)} \setminus \frac{{2}^{x}}{x}$

Proof of Result 1:
Let $\setminus \quad y = {a}^{x} \setminus q \quad \implies \setminus \ln \left(y\right) = x \setminus \ln \left(a\right)$.
Differentiating once,
$\frac{1}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \ln \left(a\right) \setminus q \quad \implies \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = y \setminus \ln \left(a\right) = {a}^{x} \setminus \ln \left(a\right)$

Proof of Result 2:
Let $\setminus \quad y = \setminus {\log}_{b} \left({x}^{n}\right)$
Using the logarithm-base rule, $\setminus {\log}_{b} \left({x}^{n}\right) = \setminus {\log}_{e} \frac{{x}^{n}}{\setminus} {\log}_{e} \left(b\right)$, we can write $y$ as -

$y = \setminus \frac{\setminus \ln \left({x}^{n}\right)}{\setminus \ln \left(b\right)} = \setminus \frac{n}{\setminus \ln \left(b\right)} \setminus \ln \left(x\right)$

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{n}{\setminus \ln \left(b\right)} \frac{1}{x} = \setminus \frac{n}{x \setminus \ln \left(b\right)}$