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Question

Differentiate ex+exexex with respect to x.

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Solution

quotient rule: ddx(uv)=vdudxudvdxv2
by chain rule: ddxeax=eaxddx(ax)=a.eax
ddxex=ex
dYdx
=(ex+ex)ddx(exex)(exex)ddx(ex+ex)(ex+ex)2
=(ex+ex)2(exex)2(ex+ex)2
=4(ex+ex)2
Y=tanhx
dYdx=sech2x=4(ex+ex)2


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