# Working with derivatives and tangents

1.) Find the derivative of the function:

a.) f(x) = x + 1/x^2

b.) f(x) = (2/3rd root of x) + 3 cos x

2.) Find equation of tangent line to the graph of f at the indicated point:

a.) y = (x^2 + 2x)(x + 1) ; (1,6)

https://brainmass.com/math/derivatives/working-derivatives-tangents-7763

## SOLUTION This solution is **FREE** courtesy of BrainMass!

1) y = x + (1/x^2)

dy/dx = 1 + d/dx(x^-2)

= 1 + [-2 * x^-3]

= 1 - (2/x^3) Answer

2)y = [2/x^(1/3)] + 3 Cos(x)

dy/dx = 2*(-1/3)[x^(-1/3)-1] + 3*-Sin(x)

= -(2/3)[1/x^(4/3)] - 3 Sin(x)

remember to include all brackets when you copy it

we used the relations

dx^n = n x^n-1

and d(cosx) = -Sinx

2) We will find the slope of the tangent at the given point first

slope = dy/dx

dy/dx = [d(x^2+2x)](x+1) + (x^2+2x)[d(x+1)] Product rule

= (2x+2)(x+1) + (x^2+2x)

Now at the point x=1 the slope is

=(2+2)(2)+ (4+4) = 8+8 = 16

Once we have a slope and a point, we can write an equation for a line using the point slope form for a line with slope m at (a,b)

(y-b) = m(x-a)

Our equation to the tangent is

y-6 =16(x-1)

or, y = 16 x -16 + 6 = 16x - 10

Please verify the steps.

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