Question

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

Answer 1

`(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`