Maths tangent question help

a curve has the equation y = x^(3) + sqrootx + (x^(2)+4x / x^(2) )

a) Find the equation of the tangent to the curve at the point where x=1

b) this tangent cuts the x axis at the point A and cuts the y axis at the point B. Find the area of the triangle OAB, where O is the origin

The equation for a tangent line is (y-b)=m(x-a) where:

m = slope of tangent line
b = value of y that that touches the line
a = value of x that touches the line

so if you work out those three variables, you can construct the equation for the tangent line

the slope of a line at any point is the derivative of the function

first lets try simplify the function, first, remember that sqrt(x) can be written as x^1/2 (for simpler application of the chain rule)

x^3 + x^1/2 + (x^2+4x/x^2)

for the chain rule, you just give the derivative of each 'added' term, and add them together, the first two terms are easy using the chain rule, where

(remember the chain rule is multiply by the power, then decrease the power by 1)

x^3 -> 3x^2
x^1/2 -> (1/2)x^-1/2

as for the fraction (x^2+4x/x^2) you could approach it a few ways, the quotient rule is the most obvious (if you remember it)

the quotient rule is: (u/v)' = (u'v -v'u)/v^2

[where ' means derivative of the term]

so if;
u = x^2 + 4x
v = x^2

then using the chain rule again

u' = 2x + 4
v' = 2x

now substitute all those terms into the equation (u'v -v'u)/v^2 to get:

[(2x + 4)(x^2) - (2x)(x^2 + 4x)] / (x^2)^2

multiply out the brackets

[(2x^3 + 4x^2) - (2x^3 + 8x^2)] / x^4

(2x^3 + 4x^2 - 2x^3 - 8x^2) / x^4

simplify

-4x^2 / x^4

-4/x^2

^ is as simple as the fraction can get, now lets add that term to the first three terms, giving the slope of the function for any given value of x