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Question: The differentiation of the given function, \( \dfrac{d}{{dx}}\left( {{a^x}} \right) \) equal to: \...

The differentiation of the given function, ddx(ax)\dfrac{d}{{dx}}\left( {{a^x}} \right) equal to:

A.x ax−1 B.ax/log⁡ae C.ax log⁡ae D.None  of  these  A.x\,{a^{x - 1}} \\\ B.{a^x}/{\log _a}e \\\ C.{a^x}\,{\log _a}e \\\ D.None\,\,of\,\,these \\\
Explanation

Solution

Hint : To find derivatives of those terms which are having variable in power. For these we first let them equal to ‘y’ and then taking log on both side and differentiating both side using formulas of derivatives to find value of dydx\dfrac{{dy}}{{dx}} or required solution of the given problem.
Formula used:
ddx(log⁡ex)=1x ddx(constan⁡t)=0 ddx(x)=1  \dfrac{d}{{dx}}\left( {{{\log }_e}x} \right) = \dfrac{1}{x} \\\ \dfrac{d}{{dx}}\left( {cons\tan t} \right) = 0 \\\ \dfrac{d}{{dx}}(x) = 1 \\\

Complete step-by-step answer :
It is required to find the derivative of ax{a^x} .
Since, variables are there in power. Therefore we first let it be equal to ‘y’.
Let, y=axy = {a^x}
Taking log on both sides of the above equation. We have
{\log _e}y = {\log _e}\left( {{a^x}} \right) \\\ \Rightarrow {\log _e}y = x.{\log _e}(a)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\\{ {{{\log }_e}{{(x)}^m} = m{{\log }_e}(x)} \right\\} \;
Now, differentiating above equation with respect to ‘x’ we have
\dfrac{d}{{dx}}\left( {{{\log }_e}y} \right) = \dfrac{d}{{dx}}\left\\{ {x.{{\log }_e}a} \right\\}
Using product rule of derivative on right hand side of above equation. We have,
\dfrac{d}{{dx}}\left( {{{\log }_e}y} \right) = x\dfrac{d}{{dx}}({\log _e}a) + \log a\dfrac{d}{{dx}}(x)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\\{ {\because \dfrac{d}{{dx}}(a) = 0} \right\\} \\\ \Rightarrow \dfrac{1}{y}\dfrac{d}{{dx}}(y) = x\left( 0 \right) + {\log _e}a\dfrac{d}{{dx}}(x) \\\ \Rightarrow \dfrac{1}{y}\dfrac{d}{{dx}}(y) = 0 + {\log _e}a\left( 1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\\{ {\because \dfrac{d}{{dx}}(x) = 1} \right\\} \\\
⇒1ydydx=log⁡ea ⇒dydx=y.log⁡ea  \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = {\log _e}a \\\ \Rightarrow \dfrac{{dy}}{{dx}} = y.{\log _e}a \\\
Substituting, value of y in above we have,
dydx=ax.log⁡e(a)\dfrac{{dy}}{{dx}} = {a^x}.{\log _e}(a)
So, the correct answer is “Option D”.

Note : For this type of derivatives problems in which options are given we can find the answer of the problem by doing integration of options given in the problem. Integration of which option ends with ax{a^x} . That will be the answer to the given problem.