Question

$\color{purple}{ \boxed{ \boxed{ \begin{matrix}\rm \red{Given \: that:-} \\ \rm \pink{ \frac{d}{dx} ({e}^{x} ) = {e}^{x} } \\ \rm \color{blue}then \: prove \: that \\ \color{green} \rm \frac{d}{dx} ({a}^x ) = {a}^{x}\ln(a) \end{matrix}}}}$​

Answer 1

$\large\underline{\sf{Solution-}}$

As it is given that,

$\rm \: \dfrac{d}{dx} ({e}^{x} ) = {e}^{x}$

Now, Consider

$\rm \: \dfrac{d}{dx}{a}^{x}$

can be rewritten as

$\rm \: = \: \dfrac{d}{dx} {e}^{log{a}^{x}}$

can be further rewritten as

$\rm \: = \: \dfrac{d}{dx} {e}^{ \: x \: loga}$

$\rm \: = \: {e}^{ \: x \: loga}\dfrac{d}{dx}(x \: loga)$

$\rm \: = \: {e}^{log{a}^{x}} \: loga \: \dfrac{d}{dx}(x)$

$\rm \: = \: {a}^{x} \: loga \: \times \: 1$

$\rm \: = \: {a}^{x} \: loga \:$

$\color{green}\rm\implies \:\boxed{ \rm{ \:\dfrac{d}{dx}{a}^{x} = \: {a}^{x} \: loga \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {x}^{n}\\ \\ \sf {nx}^{n - 1} & \sf {e}^{x} \end{array}} \\ \end{gathered}$