g(x) = 7x (2/(x^3) - 5 sq root x)?

Question

Use the rules of differentiation to find the derivatives of the following function. Simplify terms wherever possible, but it is not necessary to factorise and then cancel any of the expressions.

PLEASE GIV FULL WORKING

Answer 1

g(x) = 14 x / x ³ - 35 x^( 3 / 2 )
g(x) = 14 x^( - 2 ) - 35 x^( 3 / 2 )
g `(x) = ( - 28 ) x^(- 3) - (105 / 2) x^(1 / 2)
g `(x) = (- 28) / x ³ - (105 / 2) √x

Answer 2

g(x)=7x (2/(x^3) - 5 sq root x)

we distribute 7x factor hence

g(x)= 14 x / (x^3) - 35 x sq root x

but sq root x = x^(1/2) hence

g(x)= 14 / (x^2) - 35 x^(3/2)
= 14 x^-2 -35 x^(3/2)

We then use the following derivative rules:
1/ f(x) = a(x)+b(x) hence f'(x) = a'(x)+b'(x)
2/ f(x) = Cx^B hence f'(x) = CB x^(B-1)

hence

g'(x)= 14*(-2)* x^(-2-1) -35*(3/2) x^((3/2)-1)

hence

g'(x)= -28 * x^-3 - 105/2*x^(1/2)

and we simplify to get

g'(x)= -28 /(x^3) -52.5 sq root x