Derivative of an exponential when the exponent is a function

**jashansinghal** · July 11th 2009, 02:54 AM

what is the value of
$<br /> \frac{dy}{dx}<br />$
if
$<br /> y=2^{\sin (2^{\sin(2^x)})}<br />$

**Twig** · July 11th 2009, 03:29 AM

Use the fact that $a^{f(x)}=e^{ln(a)^{f(x)}}=e^{f(x)\cdot ln(a)}$

**Random Variable** · July 11th 2009, 03:34 AM

$y = 2^{\sin 2^{sin2^{x}}}$

let $u= 2^{x}$

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

let $v = \sin u$

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

let $w = 2^v$

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

finally let $z = \sin w$

then $y = 2^{z}$

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

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

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

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

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

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