Question

Differentiate ${2^{2x}}$.

Answer 1

Hint:We know the expression: Exponent rule:${a^b} = {e^{b\ln \left( a \right)}}$. Also we know Chain Rule:
$\dfrac{{df}}{{dx}} = \dfrac{{df}}{{du}} \times \dfrac{{du}}{{dx}}$
By using Chain rule we can solve this problem. Since we cannot find the direct derivative of the given question we have to use the chain rule. So we must convert our question in the form of the equation above such that we have to find the values of every term in the above equation and substitute it back. So by using the above information I would be able to find the solution for the given question.

Complete step by step answer:
Given, ${2^{2x}}.....................................\left( i \right)$. Here we need to find the derivative of ${2^{2x}}$.So first let’s apply the exponential rule, such that we can write ${2^{2x}}$ as:
\[{a^b} = {e^{b\ln \left( a \right)}} \\
\Rightarrow {2^{2x}} = {e^{2x\ln \left( 2 \right)}}................................\left( {ii} \right) \\ \]
So we need to find $\dfrac{d}{{dx}}{e^{2x\ln \left( 2 \right)}}$.
Now let’s apply Chain Rule:
\[{\text{So here }}f\left( x \right) = {e^{2x\ln \left( 2 \right)}} \\
\Rightarrow{\text{Let }}u = 2x\ln \left( 2 \right)...................................\left( {iii} \right) \\
\Rightarrow f\left( x \right) = {e^u}..........................................\left( {iv} \right) \\ \]
Now for applying the chain rule we need to find $\dfrac{{df}}{{du}}\;{\text{and}}\;\dfrac{{du}}{{dx}}$,
From (iii) and (iv), we can write:
$\dfrac{{df}}{{du}} = \dfrac{d}{{du}}{e^u} = {e^u} \\
\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}2x\ln \left( 2 \right) = 2\ln \left( 2 \right) \\ $
Now substituting all the values in the equation of chain rule:
$
\dfrac{{df}}{{dx}} = \dfrac{{df}}{{du}} \times \dfrac{{du}}{{dx}} \\
\Rightarrow{e^u} \times 2\ln \left( 2 \right).........................\left( v \right) \\ $
Substituting back $u$:
${e^u} \times 2\ln \left( 2 \right) = {e^{2x\ln \left( 2 \right)}} \times 2\ln \left( 2 \right)...........................\left( {vi} \right)$
So substituting in (v) we can write:
$\dfrac{{df}}{{dx}} = \dfrac{d}{{dx}}{e^{2x\ln \left( 2 \right)}} = {e^{2x\ln \left( 2 \right)}} \times 2\ln \left( 2 \right)$
Here we can write by product rule of logarithmic functions:
${e^{2x\ln \left( 2 \right)}} = {e^{\ln \left( {{2^{2x}}} \right)}}$
On simplifying we can write:
$\dfrac{d}{{dx}}{e^{2x\ln \left( 2 \right)}} = {e^{2x\ln \left( 2 \right)}} \times 2\ln \left( 2 \right) \\
\Rightarrow{e^{\ln \left( {{2^{2x}}} \right)}} \times 2\ln \left( 2 \right) \\
\Rightarrow{2^{2x}} \times 2\ln \left( 2 \right) \\
\therefore{2^{2x + 1}}\ln \left( 2 \right)................................\left( {vii} \right) \\ $
Therefore on differentiating ${2^{2x}}$ we get${2^{2x + 1}}\ln \left( 2 \right)$.

Note:Chain rule is mainly used for finding the derivative of a composite function. Here basic knowledge of exponential properties must also be known. Also care must be taken while using chain rule since it should be applied only on composite functions and applying chain rule that isn’t composite may result in a wrong derivative.