How do you use the chain rule to differentiate $2^(8x^3+8)$ ?

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Answer 1 · 2016-10-20T06:39:28

$=2^{8x^3+11}cdot 3x^2ln (2)$

$frac{d}{dx}(2^{8x^3+8})$

Applying exponential rule,

$a^b=e^{bln (a)}$

$=frac{d}{dx}(e^{(8x^3+8)ln (2)})$

$frac{df(u)}{dx}=frac{df}{du}cdot frac{du}{dx}$

Let $(8x^3+8)ln (2)=u$ ,

$=frac{d}{du}(e^u)frac{d}{dx}((8x^3+8)ln (2))$

we know,

= $frac{d}{du}(e^u)=e^u$

and,

= $frac{d}{dx}((8x^3+8)ln (2))=24x^2ln(2)$

Substituting back, $u=(8x^3+8)ln (2)$

= $e^{(8x^3+8)ln (2)}cdot 24x^2ln (2)$

Simplifying,
$=2^{8x^3+11}cdot 3x^2ln (2)$

How do you use the chain rule to differentiate 2^(8x^3+8)2^(8x^3+8)?