Please help with differentiation?

Question

The question says Given that y = a / x + 2x^2/3 and dy/dx = 7 when x =4 find the value of constant a?

Answer 1

y = (a/x) + (2x^2)/3

Let's change this around by pulling out constants from each term.

y = a(1/x) + (2/3)(x^2)

Differentiating, we have

dy/dx = a(-1/x^2) + (2/3)(2x)

Plugging in dy/dx = 7, and x = 4, we have

7 = a(-1/16) + (2/3)(8)
7 = a(-1/16) + 16/3)
7 = -a/16 + 16/3

Bringing the 16/3 over to the left hand side,

7 - 16/3 = -a/16

21/3 - 16/3 = -a/16
5/3 = -a/16

Cross multiplying,

5(16) = -a(3)
80 = -a(3), therefore
a = -80/3

Answer 2

y = a / x + (2x^2)/3

dy/dx
= -a/x^2+(4/3)x.........(1)

we are given the initial
conditions dy/dx=7,x=4

substitute into (1)

7=-a/16+(4/3)*4

a/16=(4/3)*4-7
a=16{(4/3*4-7}
=16{16/3-7}=16{-5/3}
=-80/3

therefore, a= -80/3

i hope that this helps

Answer 3

dy/dx=-a/x^2+4x/3 then we have (-a/16+16/3 = 7) then -a/16=5/3
then a = -80/3

Answer 4

(a/x)' = - a / x^2
(2x^2 / 3)' = -(2x^2)' / 3 = 4x / 3
y'(x) = - a / x^2 + 4x / 3
y'(4) = -a / 16 + 16 / 3 = 7
=> a = -80 / 3