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How do you differentiate #A=sqrt3/4s^2+(3pi)/8s^2#?

1 Answer

Apr 25, 2018

#(2sqrt3)/4s+(6pi)/8s#

Explanation:

We got: #A=sqrt3/4s^2+(3pi)/8s^2#

Differentiating respect to #s#, we get:

#(dA)/(ds)=sqrt3/4(s^2)'+(3pi)/8(s^2)'#

#=(sqrt3)/4*2s+(3pi)/8*2s#

#=(2sqrt3)/4s+(6pi)/8s#

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