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)#