Differentiate problem?

Question

Differentiate the following function.

y=(3x^2+8x+2)/sqrt(x)

I am lost at this one. I am trying to use f(x-h)-f(x)/h to solve it but keep getting the incorrect answer. any help would be great. Thanks

Answer 1

y = (3 x ² + 8 x + 2) / x^(1/2)
Use quotient rule:-
N = (x^1/2)(6x + 8) - (3x² + 8x + 2)(1/2) x^(-1/2)
N = (x^(-1/2))[(x)(6x + 8) - (3x ² + 8x + 2) ]
N = x^(-1/2) [ 3x ² - 2 ]
D = x
dy / dx = N / D
dy / dx = [ 3 x ² - 2 ] / x^(3/2)

Answer 2

This is a very simple differentiation problem.
You have to use the formula d/dx [ x^n ] = n*x^(n-1)

y = (3x^2+8x+2)/sqrt(x)
= 3*x^ (3/2) + 8*x^(1/2) + 2*x^(- 1/2)

dy/dx = 3*(3/2)*x^(1/2) + 8*(1/2)*x^( -1/2) + 2*(- 1/2)*x^( - 3/2)
= (9/2)*x^(1/2) + 4*x^( - 1/2) - x^( - 3/2)

Answer 3

It's f(x+h)-f(x)/h not f(x-h)-f(x)/h. Either way this is too tough to use first principles. Using first principles it can be shown that d/dx(x^n)=nx^(n-1) where n is any real number then you'll have to use the quotient rule.
let u=3x^2+8x+2 du/dx=6x+8, and v=sqrt(x) dv/dx=-1/2sqrt(x)
Then f'=[vu'-uv']/v^2
f'=[(6x+8)sqrt(x)+(3x^2+8x+2)/2sqrt(x)]/x

Answer 4

It's a simple quotient of functions.
f(x) = g(x)/h(x) so the derivative would be
f'(x) = (g'(x)h(x) - g(x)h'(x))/(h(x))²
If you're trying to get there with the definition of the derivative, you're doing it the hard way and leaving the door -way- too far open for little mistakes to creep in ☺

Doug