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This differentiation requires use of the chain rule. The first step is to differentiate the whole thing, treating the bracket as u, so u=1+2x². Therefore we are differentiating u^1/2....

2kx + 6ky + 4kz

=2k (x + 3y + 2z)

Since 1/5 of the class is absent, this must mean that 4/5 of the class is present. We are told that 24 pupils are present, so 4/5 of the class must equal 24 pupils. Now we need to find out how many pupils...

6a + b = 16    Equation 1
5a - 2b = 19    Equation 2

Scale up one equation so both equations have same coeffient of one variable ie 2 times Equ 1

12a + 2b = 32
5a - 2b = 19

By factorising the expression, (x-5)(x-4)=0. This means that either (x-5)=0 or (x-4)=0. By solving these equations we can work out the solutions to be x=4,5.