# Derivative of an exponential when the exponent is a function

• Jul 11th 2009, 02:54 AM
jashansinghal
Derivative of an exponential when the exponent is a function
what is the value of
$\displaystyle \frac{dy}{dx}$
if
$\displaystyle y=2^{\sin (2^{\sin(2^x)})}$
• Jul 11th 2009, 03:29 AM
Twig
Use the fact that $\displaystyle a^{f(x)}=e^{ln(a)^{f(x)}}=e^{f(x)\cdot ln(a)}$
• Jul 11th 2009, 03:34 AM
Random Variable
$\displaystyle y = 2^{\sin 2^{sin2^{x}}}$

let $\displaystyle u= 2^{x}$

then $\displaystyle y = 2^{\sin 2^{\sin u}}$

let $\displaystyle v = \sin u$

then $\displaystyle y = 2^{\sin 2^{v}}$

let $\displaystyle w = 2^v$

then $\displaystyle y = 2^{\sin w}$

finally let $\displaystyle z = \sin w$

then $\displaystyle y = 2^{z}$

$\displaystyle \frac {dy}{dx} = \frac {dy}{dz} \frac {dz}{dw} \frac {dw}{dv} \frac {dv}{du} \frac {du}{dx}$

$\displaystyle \frac {dy}{dx} = \ln(2) 2^{z} \cos(w) ln(2)2^{v} \cos(u) ln(2) 2^x$

$\displaystyle = \ln(2) 2^{\sin w} \cos(2^{v}) ln(2)2^{\sin u} \cos(2^{x}) \ln(2) 2^{x}$

$\displaystyle = \ln(2) 2^{\sin 2^{v}} \cos(2^{\sin u}) \ln(2)2^{\sin 2^{x}} \cos(2^{x}) \ln(2) 2^{x}$

$\displaystyle = \ln(2) 2^{\sin 2^{\sin 2^x}} \cos(2^{\sin 2^{x}}) \ln(2)2^{\sin 2^{x}} \cos(2^{x}) \ln(2) 2^{x}$