Question

Question #e9697

Answer 1

`0`

Explanation:

`f(x) = 6sqrt4 + x = 12+x`

`f'(x) = 1`

`f''(x) = 0`

Answer 2

0

Explanation:

Some basics
1. Derivative of constant is zero
2. Derivartive of '`(x^n)`' is `nx^(n-1)`
So solving,
`d(6*4^(1/2)+x)/dx`
so,
=1
Now diffrentiation of 1 is zero

Answer 3

`y=6sqrt(4+x) .rArr (d^2y)/dx^2=(-3/2)(x+4)^(-3/2).`

Explanation:

Let, `y=6sqrt(4+x).`

`:. dy/dx=d/dx{6(x+4)^(1/2)},`

`=6d/dx{(x+4)^(1/2)}.`

Now, we know that, `d/dx{x^n}=nx^(n-1).`Hence,

`dy/dx=6*1/2(x+4)^(1/2-1)d/dx{(x+4)},......."[The Chain Rule],"`

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

Hence, `(d^2y)/dx^2=d/dx{dy/dx},............"[Definition],"`

`=d/dx{3(x+4)^(-1/2)},`

`=3d/dx{(x+4)^(-1/2)},`

`=3{(-1/2)(x+4)^(-1/2-1)}d/dx(x+4),`

`rArr (d^2y)/dx^2=(-3/2)(x+4)^(-3/2).`