What is the derivative of #2^(2x)#?

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Answer 1 · 2016-06-05T11:58:23

#=2^{2x+1}ln (2)#

#frac{d}{dx}(2^{2x})#
Applying exponent rule,#a^b=e^{bln (a)}#

#2^{2x}=e^{2xln (2)}#
#=frac{d}{dx}(e^{2x\ln (2)})#

Applying chain rule,
#frac{df(u)}{dx}=frac{df}{du}cdot frac{du}{dx}#

Let, #2xln (2)=u#
#=frac{d}{du}(e^u)frac{d}{dx}(2xln (2))#

We know,
#frac{d}{du}(e^u)=e^u#
and,
#frac{d}{dx}(2xln (2))=2ln (2)#

Also,
#=e^u2ln (2)#

Substituting back,#u=2xln (2)#

Simplifying it,
#=2^{2x+1}ln (2)#