# Let f(x)= 6x/x-8. Then f'(15)=?

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### 3 Answers

You need to find `f'(x)` , hence, you need to differentiate the function with respect to `x ` , using quotient rule, such that:

`f'(x) = ((6x)'(x - 8) - 6x*(x - 8)')/((x - 8)^2)`

`f'(x) = (6(x - 8) - 6x*1)/((x - 8)^2)`

`f'(x) = (6x - 48 - 6x)/((x - 8)^2)`

Reducing duplicate terms yields:

`f'(x) = (-48)/((x - 8)^2)`

You need to evaluate f'(15), hence, you need to substitute 15 for x in equation `f'(x) = (-48)/((x - 8)^2),` such that:

`f'(15) = (-48)/((15 - 8)^2) => f'(15) = (-48)/(7^2)`

`f'(15) = (-48)/(49)`

**Hence, evaluating `f'(15) ` yields **`f'(15) = (-48)/(49).`

Find `f'(x)` first by applying the quotient rule:

`f'(u/v) = (vdu - udv) / v^2`

``Let u = 6x du = 6

v = x-8 dv = 1

Plug-in the given in the formula, you have:

`f'(x) = ((x-8)(6) - 6x(1))/(x-8)^(2)`

` ` `f'(x) = (6x - 48 - 6x)/(x-8)^2`

`f(x) = -48/(x-8)^2`

Evaluate the derivative when x = 15.

`f'(15) = -48/(15-8)^2`

`f'(15) = -48/(7^2)`

Therefore, `f'(15) = -48/49.`

` `

f(x)=6x/x-8

f'(x)=(v.u'-u.v')/v^2

where u=6x and v=x-8

u=6x and u'=6

v=x-8 and v'=1

f'(x)=(6(x-8)-1(6x))/(x-8)^2

(a+b)^2=(a^2+2ab+b^2)

(a-b)^2=(a^2-2ab+b^2)

f'(x)=(6x-48-6x)/(x^2-16x+64)

f'(x)=-48/(x^2-16x+64)

f'(15)=-48/(15^2-16(15)+64)

f'(15)=-48/49` ` ` `