Question

heya!! ...❤❤ur question is here! !..
content quality required. .....any genius! !!...

show that -: differentiation for a to the power x is equal to a to the power x log a ....see attachment

Answer 1

Hey friend, Harish here.

Here is your answer.

To Show:

$\frac{d}{dx} (a^x) = a^x \ln (a)$

Solution:

We know that ;

⇒ $a^x = e^{x \ln (a)}$

Then ;

⇒ $\frac{d}{dx}( e^{x \ln (a)})$  

Let c = x ln (a) 

So, By applying chain rule :

⇒ $\frac{d}{dx}( e^{x \ln (a)}) = \frac{d}{dc} (e^c) . \frac{d}{dx}(x \ln (a))$

⇒ $e^{x \ln (a)} . \ln (a) = a^x \ln (a)$
_________________________________________________

Hope my answer is helpful to you.

Answer 2

$\color{gold}{\bf \: HELLO \: \:DEAR,}$

$\bf let \: y = a^x$

$\color{skyblue}{\bf {taking \: \: log \: \: both \: \: side,}} \\ \\ \bf log_ey = log_ea^x \\ \\ \bf log_ey = xlog_ea \\ \\ \bf 1/y * dy/dx = x * d(log_ea)/dx + log_ea * da/dx \\ \\ \bf dy/dx = y * log_ea \\ \\ \bf dy/dx = a^x log_ea$

$\color{red}\underline{\bf I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}$